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Motion PYQs – Physics, SSC & Railway

123 Motion (Physics) questions asked in recent SSC and Railway exams, newest first, with exam, date & shift, answers and solutions.

📚 123 questionsSSC 12 · Railway 111
12
RRB Group D 2025-26 – 25 August 2026 Shift 2 · Speed and velocity

Q1. How does velocity differ from speed in the context of describing motion?

  1. (a) Velocity specifies both the magnitude (speed) and the definite direction of motion.
  2. (b) Velocity applies only to non-uniform motion, while speed applies only to uniform motion.
  3. (c) Velocity is calculated using average speed, while speed uses instantaneous speed.
  4. (d) Velocity specifies distance covered, while speed specifies displacement.
Answer & solution

Answer: (a) Velocity specifies both the magnitude (speed) and the definite direction of motion.

Correct Answer:
Correct Answer: Option 1 – Velocity specifies both the magnitude (speed) and the definite direction of motion.
Explanation:
Speedis a scalar quantity — it tells only how fast an object moves (magnitude).Velocityis a vector quantity — it specifies both the speed (magnitude) and the direction of motion. This is the key conceptual difference between the two.
RRB Group D 2025-26 – 25 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q2. A student observes a speed-time graph for a car that moves with constant speed for 10 seconds, then accelerates for 5 seconds. How can the student use the graph to calculate the total distance travelled during these 15 seconds?

  1. (a) By using only the area under the acceleration section
  2. (b) By calculating the area under the speed-time graph for each section and adding them
  3. (c) By multiplying the total time by the average speed shown on the graph
  4. (d) By reading the final speed at 15 seconds and multiplying by 15
Answer & solution

Answer: (b) By calculating the area under the speed-time graph for each section and adding them

Correct Answer:
Correct Answer: Option 2 – By calculating the area under the speed-time graph for each section and adding them
Explanation:
The area under a speed-time graph gives distance travelled. Since the motion has two distinct sections (constant speed, then acceleration), the student should compute the area of each section separately (a rectangle for constant speed, a trapezoid for the accelerating part) and sum them to get total distance.
Important Facts:
• Area under a v-t graph = displacement/distance.
• For uniform motion the area is a rectangle; for uniform acceleration it is a trapezoid or triangle plus rectangle.
• Reading just the final speed or using only one section's area gives incorrect (incomplete) total distance.
RRB Group D 2025-26 – 21 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q3. Which of the following quantities does the slope of a velocity-time graph represent for an object in motion?

  1. (a) Displacement
  2. (b) Force
  3. (c) Acceleration
  4. (d) Speed
Answer & solution

Answer: (c) Acceleration

Correct Answer:
Correct Answer: Option 3 – Acceleration
Explanation:
On a velocity-time graph, slope = change in velocity / change in time = acceleration. The area under the curve, not the slope, gives displacement.
RRB Group D 2025-26 – 21 August 2026 Shift 2 · Uniform circular motion

Q4. Which statement best describes the direction of velocity in uniform circular motion?

  1. (a) It is always along the radius away from the center
  2. (b) It is always tangent to the circle at any point
  3. (c) It is always perpendicular to the plane of the circle
  4. (d) It is always directed towards the center
Answer & solution

Answer: (b) It is always tangent to the circle at any point

Correct Answer:
Correct Answer: Option 2 – It is always tangent to the circle at any point
Explanation:
In uniform circular motion, the velocity vector at any instant is directed along the tangent to the circular path at that point, perpendicular to the radius. It is the centripetal acceleration (and net force) that is directed towards the centre, not the velocity.
RRB Group D 2025-26 – 21 August 2026 Shift 3 · Numericals on motion

Q5. A cyclist travels 60 metres north and then 80 metres south in 20 seconds. What is the magnitude of his average velocity?

  1. (a) Five metres per second
  2. (b) One metre per second
  3. (c) Zero metres per second
  4. (d) Seven metres per second
Answer & solution

Answer: (b) One metre per second

Correct Answer:
Correct Answer: Option 2 – One metre per second
Explanation:
Displacement (net) = 80 m south − 60 m north =20 m(net direction south). Average velocity = displacement ÷ time = 20 m ÷ 20 s =1 m/s.
Important Facts:
• Average velocity is calculated using net displacement, not total distance travelled.
• Average speed here would instead use total distance (60+80=140 m), giving 7 m/s — a common distractor (Option 4).
RRB Group D 2025-26 – 20 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q6. Which method is appropriate for finding the distance travelled by an object when its velocity-time graph is a straight line sloping upwards from the origin?

  1. (a) Read the final velocity value from the graph
  2. (b) Calculate the slope of the velocity-time graph
  3. (c) Subtract the initial time from the final time
  4. (d) Calculate the area under the velocity-time graph for the given interval
Answer & solution

Answer: (d) Calculate the area under the velocity-time graph for the given interval

Correct Answer:
Correct Answer: Option 4 – Calculate the area under the velocity-time graph for the given interval
Explanation:
On a velocity-time graph, thearea under the graph(between the line and the time axis, over the given interval) gives thedistance/displacementtravelled by the object. For a straight line through the origin, this area is a triangle whose area equals ½ × base × height.
Important Facts:
• The slope of a velocity-time graph gives acceleration, not distance.
• For uniformly accelerated motion starting from rest, area under v-t graph = ½ × t × v = distance travelled.
RRB Group D 2025-26 – 20 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q7. Which of the following best describes the slope of the velocity-time graph for uniform velocity?

  1. (a) The slope changes at every point
  2. (b) The slope is negative and constant
  3. (c) The slope is positive and constant
  4. (d) The slope is zero at all points
Answer & solution

Answer: (d) The slope is zero at all points

Correct Answer:
Correct Answer: Option 4 – The slope is zero at all points.
Explanation:
On a velocity–time graph, the slope representsacceleration. For uniform (constant) velocity, acceleration is zero throughout the motion, so the v–t graph is a horizontal straight line with zero slope.
Important Facts:
• A positive constant slope on a v-t graph represents uniform acceleration, not uniform velocity.
• The area under a v-t graph gives displacement.
RRB Group D 2025-26 – 20 August 2026 Shift 2 · Numericals on motion

Q8. If a car moves with a speed of 10 m/s for 5 seconds, what distance does it cover?

  1. (a) 50 meters
  2. (b) 5 meters
  3. (c) 100 meters
  4. (d) 15 meters
Answer & solution

Answer: (a) 50 meters

Correct Answer:
Correct Answer: Option 1 – 50 meters
Explanation:
Distance = Speed × Time = 10 m/s × 5 s = 50 m.
RRB Group D 2025-26 – 19 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q9. If the velocity-time graph of a moving object shows a straight line parallel to the time axis, what can be concluded about the object's acceleration?

  1. (a) The object has negative acceleration
  2. (b) The object has uniform acceleration
  3. (c) The object has zero acceleration
  4. (d) The object has changing acceleration
Answer & solution

Answer: (c) The object has zero acceleration

Correct Answer:
Correct Answer: Option 3 – The object has zero acceleration
Explanation:
A straight lineparallel to the time axison a velocity-time graph means the velocity staysconstantover time. Since acceleration is the rate of change of velocity, a constant velocity implieszero acceleration.
Important Facts:
• Slope of a v-t graph gives acceleration.
• A straight line with a slope indicates uniform (non-zero) acceleration; a horizontal line indicates zero acceleration.
RRB Group D 2025-26 – 19 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q10. Which quantity is represented by the slope of a position-time graph for an object moving in a straight line?

  1. (a) Momentum
  2. (b) Linear velocity
  3. (c) Acceleration
  4. (d) Force
Answer & solution

Answer: (b) Linear velocity

Correct Answer:
Correct Answer: Option 2 – Linear velocity
Explanation:
On a position-time (displacement-time) graph, theslopeat any point equals the rate of change of position with respect to time, which is the definition ofvelocity. A steeper slope indicates higher velocity; a horizontal line (zero slope) indicates the object is at rest.
Important Facts:
• Slope of a velocity-time graph gives acceleration, not the slope of a position-time graph.
• A curved position-time graph indicates changing velocity (i.e., acceleration is present).
RRB Group D 2025-26 – 19 August 2026 Shift 3 · Acceleration and retardation

Q11. Acceleration is defined as:

  1. (a) Displacement per unit time
  2. (b) Change in velocity per unit time
  3. (c) Distance per unit time
  4. (d) Change in displacement per unit time
Answer & solution

Answer: (b) Change in velocity per unit time

Correct Answer:
Option 2 – Change in velocity per unit time
Explanation:
Acceleration is defined as therate of change of velocity with respect to time, i.e., a = Δv/Δt. It is a vector quantity, measured in m/s².
Important Facts:
• Velocity = displacement per unit time.
• Speed = distance per unit time.
• Acceleration can be positive (speeding up) or negative/retardation (slowing down).
Correct Answer: Option 2 – Change in velocity per unit time
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Uniform circular motion

Q12. Which of the following best describes the motion of a car traveling at a constant speed around a circular track?

  1. (a) The car moves with uniform motion as its velocity remains constant in both magnitude and direction.
  2. (b) The car experiences non-uniform motion because it moves along a curved path at constant speed.
  3. (c) The car moves with uniform circular motion because its speed is constant, but its direction changes continuously.
  4. (d) The car moves in a straight line at constant speed since its direction never changes.
Answer & solution

Answer: (c) The car moves with uniform circular motion because its speed is constant, but its direction changes continuously.

Motion on a circular track
A car moving at constant speed on a circular track has a constantly changing direction of motion, even though its speed (magnitude of velocity) does not change.
Since velocity is a vector, a change in direction means velocity is changing — this is calleduniform circular motion: constant speed, continuously changing direction.
Correct Answer: The car moves with uniform circular motion because its speed is constant, but its direction changes continuously.
RRB Group D 2025-26 – 18 August 2026 Shift 1 · Numericals on motion

Q13. A runner moves 60 meters north and then 80 meters south in 20 seconds. Which statement is correct about her average speed and average velocity?

  1. (a) Average speed is greater than average velocity
  2. (b) Average speed is less than average velocity
  3. (c) Average speed is zero
  4. (d) Average speed equals average velocity
Answer & solution

Answer: (a) Average speed is greater than average velocity

Average speed vs average velocity
• Total distance travelled = 60 m + 80 m = 140 m; Time = 20 s
• Average speed = 140/20 = 7 m/s
• Net displacement = 80 m south − 60 m north = 20 m south
• Average velocity = 20/20 = 1 m/s (south)
Since 7 m/s > 1 m/s, average speed is greater than average velocity.
Correct Answer: Average speed is greater than average velocity
RRB Group D 2025-26 – 18 August 2026 Shift 2 · Numericals on motion

Q14. A car travels 150 kilometers in 3 hours. What is the average speed of the car, and what is the correct unit for this quantity?

  1. (a) 0.02 kilometers per hour
  2. (b) 150 kilometers per hour
  3. (c) 50 kilometers per minute
  4. (d) 50 kilometers per hour
Answer & solution

Answer: (d) 50 kilometers per hour

Average Speed Calculation
Average speed = Total distance / Total time = 150 km / 3 h =50 km/h.
The correct unit for speed is distance per unit time — kilometers per hour, matching option 4.
Correct Answer: 50 kilometers per hour
RRB Group D 2025-26 – 18 August 2026 Shift 3 · Uniform circular motion

Q15. A toy train moves around a circular track at a constant speed. Which statement correctly describes its motion?

  1. (a) The train moves with increasing speed along the circular track.
  2. (b) The train changes its direction continuously while moving with constant speed.
  3. (c) The train moves in a straight line with constant speed.
  4. (d) The train remains at rest because its speed does not change.
Answer & solution

Answer: (b) The train changes its direction continuously while moving with constant speed.

Concept: Uniform Circular Motion
When an object moves along a circular path at constant speed, its speed (magnitude of velocity) stays the same, but its direction of motion changes continuously at every point of the path. Since velocity is a vector (speed + direction), this constitutes accelerated motion even though the speed itself is constant — this is the defining feature of uniform circular motion.
Correct Answer: The train changes its direction continuously while moving with constant speed.
RRB Group D 2025-26 – 18 August 2026 Shift 3 · Speed and velocity

Q16. A cyclist covers 2 km in the first 10 minutes and 3 km in the next 10 minutes. What type of motion is this?

  1. (a) periodic motion
  2. (b) non-uniform motion
  3. (c) circular motion
  4. (d) uniform motion
Answer & solution

Answer: (b) non-uniform motion

Concept: Uniform vs Non-Uniform Motion
In uniform motion, an object covers equal distances in equal intervals of time. Here, the cyclist covers 2 km in the first 10 minutes but 3 km in the next 10 minutes — unequal distances in equal time intervals, meaning the speed is changing. This is the definition ofnon-uniform motion.
Correct Answer: non-uniform motion
RRB Group D 2025-26 – 17 August 2026 Shift 1 · Numericals on motion

Q17. If a person walks 5 meters east, then 12 meters west along a straight path, which value represents the magnitude of the person’s displacement?

  1. (a) 12 meters
  2. (b) 17 meters
  3. (c) 7 meters
  4. (d) 5 meters
Answer & solution

Answer: (c) 7 meters

Magnitude of Displacement
Taking east as positive: the person walks +5 m (east), then −12 m (west). Net displacement = 5 − 12 = −7 m, i.e., 7 m towards the west.
Themagnitudeof displacement (distance and direction ignored, only size) is7 meters.
Correct Answer: 7 meters
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Uniform circular motion

Q18. Which of the following best describes uniform circular motion?

  1. (a) An object moving at a constant speed in a straight line
  2. (b) An object moving at a constant speed along a circular path
  3. (c) An object at rest in a circular path
  4. (d) An object moving at a changing speed along a curved path
Answer & solution

Answer: (b) An object moving at a constant speed along a circular path

Uniform Circular Motion
Uniform circular motion is defined as motion of an object along a circular path at aconstant speed(though velocity direction keeps changing, resulting in acceleration towards the centre).
• Constant speed in a straight line is uniform linear motion, not circular.
• An object at rest is not in motion at all.
Correct Answer: An object moving at a constant speed along a circular path
RRB Group D 2025-26 – 17 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q19. If the distance between two points on a straight line distance-time graph is 120 meters and the corresponding time interval is 4 seconds, what is the uniform speed?

  1. (a) 30 m/s
  2. (b) 60 m/s
  3. (c) 24 m/s
  4. (d) 15 m/s
Answer & solution

Answer: (a) 30 m/s

Speed from Distance-Time Graph
For a straight-line distance-time graph, the slope gives the uniform speed.
• Speed = Distance / Time = 120 m / 4 s = 30 m/s
Correct Answer: 30 m/s
RRB Group D 2025-26 – 14 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q20. On a velocity-time graph, what does a horizontal straight line above the time axis indicate about the motion of the object?

  1. (a) Constant velocity
  2. (b) Increasing velocity
  3. (c) Constant acceleration
  4. (d) Changing direction
Answer & solution

Answer: (a) Constant velocity

Answer: Option 1 – Constant velocity.
• A horizontal line on a velocity–time graph means velocity does not change ⇒constant velocity.
RRB Group D 2025-26 – 14 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q21. A velocity-time graph for a body moving with uniform acceleration is a straight line that slopes upwards. What does the area under this line between two time points represent?

  1. (a) The acceleration of the body during that time interval
  2. (b) The displacement of the body during that time interval
  3. (c) The average velocity during that time interval
  4. (d) The total change in velocity during that time interval
Answer & solution

Answer: (b) The displacement of the body during that time interval

Answer: Option 2 – The displacement of the body during that time interval.
• Area under a velocity–time graph =displacement.
RRB Group D 2025-26 – 14 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q22. When interpreting a velocity–time graph, what physical quantity is represented by the area under the graph during a specific time interval?

  1. (a) Speed of the car
  2. (b) Distance (or displacement) travelled
  3. (c) Acceleration of the car
  4. (d) Change in velocity
Answer & solution

Answer: (b) Distance (or displacement) travelled

Answer: Option 2 – Distance (or displacement) travelled.
• Area under a velocity–time graph gives thedistance/displacement.
RRB Group D 2025-26 – 14 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q23. A velocity-time graph for a car moving with uniform velocity starts at t=0 s and ends at t=10 s, always at 5 m/s. What is the shape of the graph?

  1. (a) A curve rising above 5 m/s
  2. (b) A straight line rising from 0 to 5 m/s
  3. (c) A straight horizontal line at 5 m/s
  4. (d) A straight line falling from 5 to 0 m/s
Answer & solution

Answer: (c) A straight horizontal line at 5 m/s

Answer: Option 3 – A straight horizontal line at 5 m/s.
• Uniform velocity ⇒ velocity constant ⇒ ahorizontal lineon a velocity–time graph.
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q24. In a velocity–time graph, which type of line represents motion with no change in velocity?

  1. (a) Slanted straight line
  2. (b) Parabola
  3. (c) Horizontal straight line
  4. (d) Curved line
Answer & solution

Answer: (c) Horizontal straight line

Answer: Option 3 – Horizontal straight line.
• On a velocity–time graph, constant velocity (no change) is ahorizontal line.
RRB Group D 2025-26 – 13 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q25. A distance-time graph for an object moving at uniform speed is characterized by which of the following features?

  1. (a) A horizontal straight line
  2. (b) A curved line with changing slope
  3. (c) A straight line with a constant positive slope
  4. (d) A straight line with a negative slope
Answer & solution

Answer: (c) A straight line with a constant positive slope

Answer: Option 3 – A straight line with a constant positive slope.
• Uniform speed on a distance–time graph is astraight line with constant slope.
RRB Group D 2025-26 – 13 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q26. For a car moving with accelerated motion, the distance-time graph is:

  1. (a) Straight line
  2. (b) Non-linear curve
  3. (c) Vertical line
  4. (d) Horizontal line
Answer & solution

Answer: (b) Non-linear curve

Answer: Option 2 – Non-linear curve.
• For accelerated motion, the distance–time graph is acurve(non-linear).
RRB Group D 2025-26 – 13 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q27. Two objects, P and Q, move with uniform speeds. The distance–time graph of P is steeper than that of Q. What does this indicate?

  1. (a) P has lower speed than Q
  2. (b) P is accelerating
  3. (c) Both have equal speed
  4. (d) P has greater speed than Q
Answer & solution

Answer: (d) P has greater speed than Q

Answer: Option 4 – P has greater speed than Q.
• A steeper distance–time line means agreater slope (speed), so P is faster.
RRB Group D 2025-26 – 13 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q28. Which feature of a velocity-time graph for uniform acceleration determines the value of acceleration?

  1. (a) The slope of the velocity-time graph
  2. (b) The maximum value on the velocity axis
  3. (c) The area under the velocity-time graph
  4. (d) The intercept on the velocity axis
Answer & solution

Answer: (a) The slope of the velocity-time graph

Answer: Option 1 – The slope of the velocity-time graph.
• On a velocity–time graph,slope = acceleration.
RRB Group D 2025-26 – 12 August 2026 Shift 1 · Distance and displacement

Q29. Which of the following vectors correctly represents the displacement if an object moves from the point (0, 0) to the point (4, -3) in a plane?

  1. (a) (-3, 4)
  2. (b) (3, -4)
  3. (c) (4, -3)
  4. (d) (-4, 3)
Answer & solution

Answer: (c) (4, -3)

Answer: Option 3 – (4, −3).
• Displacement from (0,0) to (4,−3) is the vector(4, −3).
RRB Group D 2025-26 – 12 August 2026 Shift 2 · Speed and velocity

Q30. Two objects, Object P and Object Q, move along a straight road. Object P travels 20 meters every 2 seconds, while Object Q covers 20 meters in the first 2 seconds and 30 meters in the next 2 seconds. Which statement is correct about their motions?

  1. (a) Object P is in uniform motion, while Object Q is in non-uniform motion.
  2. (b) Both objects are in non-uniform motion.
  3. (c) Object P is in non-uniform motion, while Object Q is in uniform motion.
  4. (d) Both objects are in uniform motion.
Answer & solution

Answer: (a) Object P is in uniform motion, while Object Q is in non-uniform motion.

Answer: Option 1 – Object P is in uniform motion, while Object Q is in non-uniform motion.
• P covers equal distances (20 m) in equal times ⇒ uniform. Q covers 20 m then 30 m ⇒ non-uniform.
RRB Group D 2025-26 – 12 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q31. Which of the following situations shows an object with zero linear velocity on a position-time graph?

  1. (a) An upward-sloping straight line from the origin.
  2. (b) A downward-sloping straight line from the origin.
  3. (c) A curved line rising from left to right.
  4. (d) A horizontal straight line parallel to the time axis.
Answer & solution

Answer: (d) A horizontal straight line parallel to the time axis.

Answer: Option 4 – A horizontal straight line parallel to the time axis.
• On a position–time graph, zero velocity = zero slope = ahorizontal line(position not changing).
RRB Group D 2025-26 – 11 August 2026 Shift 2 · Acceleration and retardation

Q32. Which physical quantity is defined as the rate of change of velocity with respect to time?

  1. (a) Distance travelled
  2. (b) Speed
  3. (c) Displacement
  4. (d) Acceleration
Answer & solution

Answer: (d) Acceleration

Answer: Option 4 – Acceleration.
•Acceleration= rate of change of velocity with time.
RRB Group D 2025-26 – 10 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q33. A velocity–time graph is represented by a straight line parallel to the time axis and lying above it. What does this indicate about the motion of the object?

  1. (a) The object is moving with constant velocity
  2. (b) The object is moving with decreasing velocity
  3. (c) The object is at rest
  4. (d) The object is moving with increasing velocity
Answer & solution

Answer: (a) The object is moving with constant velocity

Answer: Option 1 — The object is moving with constant velocity
A v–t line parallel to the time axis (above it) meansconstant velocity→ Option 1.
RRB Group D 2025-26 – 10 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q34. A Distance-Time graph shows a curve that starts steep and gradually becomes flatter (slope decreasing). This motion represents __________.

  1. (a) Zero acceleration
  2. (b) Deceleration
  3. (c) Positive acceleration
  4. (d) Constant, negative velocity
Answer & solution

Answer: (b) Deceleration

Answer: Option 2 – Deceleration.
• On a distance–time graph, slope = speed. A curve that starts steep and becomes flatter means theslope (speed) is decreasing⇒ the body is slowing down =deceleration(retardation).
RRB Group D 2025-26 – 10 August 2026 Shift 3 · Distance and displacement

Q35. Which statement correctly describes the difference between position vector and displacement vector?

  1. (a) Position vector locates a point; displacement vector shows the change between two points
  2. (b) Both always have the same magnitude
  3. (c) Both always point in the same direction
  4. (d) Position vector always shows the path traveled
Answer & solution

Answer: (a) Position vector locates a point; displacement vector shows the change between two points

Answer: Option 1 – Position vector locates a point; displacement vector shows the change between two points.
• Aposition vectorgives a point's location relative to the origin; adisplacement vectoris the change in position between two points.
RRB Group D 2025-26 – 9 August 2026 Shift 1 · Distance and displacement

Q36. What does the displacement vector represent in two-dimensional motion?

  1. (a) The distance traveled along the path
  2. (b) The average speed of the object
  3. (c) The path followed by the object
  4. (d) The change in position from the initial point to the final point
Answer & solution

Answer: (d) The change in position from the initial point to the final point

Answer: Option 4 — Change in position from initial to final point
Displacement is thestraight-line change in positionfrom start to end → Option 4.
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Speed and velocity

Q37. Which quantity does linear velocity describe in the context of motion?

  1. (a) The amount of force applied
  2. (b) The rate of change of position along a straight path
  3. (c) The total distance covered in all directions
  4. (d) The direction of motion only
Answer & solution

Answer: (b) The rate of change of position along a straight path

Answer: Option 2 — The rate of change of position along a straight path
Linear velocity =rate of change of position along a line→ Option 2.
RRB Group D 2025-26 – 9 August 2026 Shift 2 · Distance and displacement

Q38. Which statement best describes the difference between path length and displacement for an object moving in a straight line and returning to its starting point?

  1. (a) Displacement is always negative
  2. (b) Path length is always equal to displacement
  3. (c) Displacement is greater than path length
  4. (d) Path length is either equal to or greater than displacement
Answer & solution

Answer: (d) Path length is either equal to or greater than displacement

Answer: Option 4 — Path length is either equal to or greater than displacement
Path length ≥ displacement always → Option 4.
RRB Group D 2025-26 – 9 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q39. Which of the following best describes the shape of a distance-time graph for an object moving at a uniform speed?

  1. (a) A curved line upward
  2. (b) A straight line with varying slope
  3. (c) A straight line with a constant slope
  4. (d) A horizontal straight line
Answer & solution

Answer: (c) A straight line with a constant slope

Answer: Option 3 — A straight line with a constant slope
Uniform speed ⟹ distance–time graph is astraight line with constant slope→ Option 3.
RRB Group D 2025-26 – 9 August 2026 Shift 3 · Numericals on motion

Q40. A car starts from rest and accelerates uniformly at 2 m/s² for 5 seconds. What is the distance it travels during this time?

  1. (a) 20 meters
  2. (b) 25 meters
  3. (c) 10 meters
  4. (d) 50 meters
Answer & solution

Answer: (b) 25 meters

Answer: Option 2 — 25 meters
s = ut + ½at² = 0 + ½ × 2 × 5² =25 m→ Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q41. Which statement best describes a straight line sloping upwards on a distance-time graph for an object moving with uniform speed?

  1. (a) The object is stationary at all times
  2. (b) The object moves with varying speed throughout
  3. (c) The object returns to its starting point repeatedly
  4. (d) The object covers equal distances in equal intervals of time
Answer & solution

Answer: (d) The object covers equal distances in equal intervals of time

Answer: Option 4 — Covers equal distances in equal intervals of time
A straight upward-sloping distance–time line meansuniform speed— equal distance in equal time → Option 4.
RRB Group D 2025-26 – 6 August 2026 Shift 1 · Uniform circular motion

Q42. Which statement best describes the velocity of an object undergoing uniform circular motion?

  1. (a) The object’s speed increases and decreases as it moves around the circle.
  2. (b) The object moves with a changing velocity along a straight path.
  3. (c) The velocity remains unchanged in both magnitude and direction throughout motion.
  4. (d) The magnitude of velocity remains constant, but its direction continuously changes around the circle.
Answer & solution

Answer: (d) The magnitude of velocity remains constant, but its direction continuously changes around the circle.

Answer: Option 4 — Magnitude constant, direction continuously changes
In uniform circular motion,speed (magnitude) is constantbut thedirection of velocity changes→ Option 4.
RRB Group D 2025-26 – 6 August 2026 Shift 2 · Numericals on motion

Q43. A car moves at a constant speed of 20 m/s for 5 seconds. What is the total distance travelled by the car during this time?

  1. (a) 25 meters
  2. (b) 100 meters
  3. (c) 400 meters
  4. (d) 4 meters
Answer & solution

Answer: (b) 100 meters

Answer: Option 2 — 100 meters
Distance = speed × time = 20 × 5 =100 m→ Option 2.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Uniform circular motion

Q44. Which quantity changes continuously for an object in uniform circular motion?

  1. (a) The direction of velocity changes
  2. (b) The speed changes
  3. (c) The mass changes
  4. (d) The kinetic energy changes
Answer & solution

Answer: (a) The direction of velocity changes

Answer: Option 1 — The direction of velocity changes
In uniform circular motion, speed and mass stay constant, but thedirection of velocity changes continuously→ Option 1.
RRB Group D 2025-26 – 6 August 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q45. Which of the following best describes the appearance of a distance-time graph for an object moving at a uniform speed?

  1. (a) A horizontal line parallel to the time axis
  2. (b) A straight line with a constant positive slope
  3. (c) A vertical line parallel to the distance axis
  4. (d) A curved line with a changing slope
Answer & solution

Answer: (b) A straight line with a constant positive slope

Answer: Option 2 — A straight line with a constant positive slope
Uniform speed ⟹ distance–time graph is astraight line with constant slope→ Option 2.
RRB Group D 2025-26 – 5 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q46. A car starts from rest and gradually increases its speed. Which of the following best describes its distance-time graph?

  1. (a) A curve starting from the origin with increasing slope
  2. (b) A straight line with constant slope starting from the origin
  3. (c) A horizontal line starting from the origin
  4. (d) A curve starting from the origin with decreasing slope
Answer & solution

Answer: (a) A curve starting from the origin with increasing slope

Answer: Option 1 — A curve starting from the origin with increasing slope
Speed increasing ⟹ distance grows faster with time ⟹ acurve with increasing slopefrom the origin → Option 1.
RRB Group D 2025-26 – 5 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q47. Which statement best describes the velocity-time graph for an object moving with uniform acceleration?

  1. (a) It is a horizontal line parallel to the time axis
  2. (b) It is a curve bending upwards
  3. (c) It is a straight line parallel to the velocity axis
  4. (d) It is a straight line with a constant slope
Answer & solution

Answer: (d) It is a straight line with a constant slope

Answer: Option 4 — A straight line with a constant slope
Uniform acceleration ⟹ velocity changes at a constant rate ⟹ v–t graph is astraight line with constant slope→ Option 4.
RRB Group D 2025-26 – 5 August 2026 Shift 2 · Numericals on motion

Q48. A car increases its velocity from 10 m/s to 30 m/s in 5 seconds. What is the acceleration of the car?

  1. (a) 6 m/s²
  2. (b) 10 m/s²
  3. (c) 2 m/s²
  4. (d) 4 m/s²
Answer & solution

Answer: (d) 4 m/s²

Answer: Option 4 — 4 m/s²
a = (v − u)/t = (30 − 10)/5 = 20/5 =4 m/s²→ Option 4.
RRB Group D 2025-26 – 5 August 2026 Shift 3 · Speed and velocity

Q49. Which of the following describes an object moving with uniform motion along a straight line?

  1. (a) An object accelerating as it moves forward
  2. (b) An object traveling at a constant speed in one direction
  3. (c) An object changing direction at intervals
  4. (d) An object stopping and starting repeatedly
Answer & solution

Answer: (b) An object traveling at a constant speed in one direction

Answer: Option 2 — Constant speed in one direction
Uniform motion along a straight line =constant speed in one direction→ Option 2.
RRB Group D 2025-26 – 4 August 2026 Shift 1 · Distance and displacement

Q50. What is required to fully describe the displacement of a moving body?

  1. (a) Position at the origin
  2. (b) Both magnitude and direction
  3. (c) Only direction
  4. (d) Only magnitude
Answer & solution

Answer: (b) Both magnitude and direction

Answer: Option 2 — Both magnitude and direction
Displacement is avector, so it needsboth magnitude and direction→ Option 2.
RRB Group D 2025-26 – 4 August 2026 Shift 2 · Distance and displacement

Q51. Which quantity can never be negative in one-dimensional motion?

  1. (a) Displacement
  2. (b) Path length
  3. (c) Velocity
  4. (d) Change in position
Answer & solution

Answer: (b) Path length

Answer: Option 2 — Path length
Path length (distance)is always ≥ 0; displacement and velocity can be negative in 1-D → Option 2.
RRB Group D 2025-26 – 4 August 2026 Shift 3 · Acceleration and retardation

Q52. Which of the following describes the rate at which an object changes its velocity?

  1. (a) Acceleration
  2. (b) Speed
  3. (c) Displacement
  4. (d) Distance
Answer & solution

Answer: (a) Acceleration

Answer: Option 1 — Acceleration
Accelerationis the rate of change of velocity → Option 1.
RRB Group D 2025-26 – 4 August 2026 Shift 3 · Uniform circular motion

Q53. Which statement best describes uniform circular motion in the context of motion and force?

  1. (a) An object moves in a straight line at changing speed.
  2. (b) An object moves in a circle at constant speed but its velocity changes due to change in direction.
  3. (c) An object moves in a circle with both constant speed and constant direction.
  4. (d) An object moves in a circle with changing speed and constant direction.
Answer & solution

Answer: (b) An object moves in a circle at constant speed but its velocity changes due to change in direction.

Answer: Option 2 — Circle at constant speed with velocity changing due to direction
In uniform circular motion,speed is constant but velocity changes(direction changes) → Option 2.
RRB Group D 2025-26 – 3 August 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q54. What does a position-time graph with a straight line parallel to the time axis represent in the context of an object’s motion?

  1. (a) The object is at rest, covering no distance over time
  2. (b) The object moves with uniform speed over time
  3. (c) The object moves with varying speed
  4. (d) The object accelerates over time
Answer & solution

Answer: (a) The object is at rest, covering no distance over time

Answer: Option 1 — The object is at rest
Concept:In a position–time (x–t) graph, theslope = speed.
A lineparallel to the time axisis horizontal → slope = 0 → speed = 0.
So the position does not change with time: the object isat rest, covering no distance.
(A straight slanted line = uniform speed; a curved line = changing speed/acceleration.)
RRB Group D 2025-26 – 3 August 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q55. From a distance-time graph, how can the uniform speed of an object be calculated?

  1. (a) By measuring the area under the graph
  2. (b) By measuring the total time only
  3. (c) By measuring the slope of the graph
  4. (d) By measuring the total distance only
Answer & solution

Answer: (c) By measuring the slope of the graph

Answer: Option 3 — By measuring the slope of the graph
In a distance–time graph,speed = slope= (change in distance)/(change in time). For uniform speed the graph is a straight line, and its slope gives the constant speed → Option 3.
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Uniform circular motion

Q56. Which of the following statements best defines uniform circular motion?

  1. (a) Motion along a circular path with constant velocity and zero acceleration.
  2. (b) Motion along a circular path with constant speed and changing direction of velocity.
  3. (c) Motion along a circular path with changing speed but constant velocity.
  4. (d) Motion along a straight path with constant speed and constant velocity.
Answer & solution

Answer: (b) Motion along a circular path with constant speed and changing direction of velocity.

Answer: Option 2 — Constant speed and changing direction of velocity
Uniform circular motion:thespeed is constantbut thedirection of velocity changes continuously, so velocity is not constant and there is (centripetal) acceleration → Option 2.
RRB Group D 2025-26 – 3 August 2026 Shift 3 · Numericals on motion

Q57. A cyclist covers a distance of 30 kilometers in 1.5 hours. What is the average speed of the cyclist?

  1. (a) 20 kilometers per hour
  2. (b) 2 kilometers per hour
  3. (c) 15 kilometers per hour
  4. (d) 45 kilometers per hour
Answer & solution

Answer: (a) 20 kilometers per hour

Answer: Option 1 — 20 km/h
Average speed = distance/time = 30/1.5 =20 km/h→ Option 1.
SSC GD 2026 – 27 April 2026 Shift 2 · Acceleration and retardation

Q58. When is acceleration considered positive?

  1. (a) When it is opposite to velocity
  2. (b) When velocity is decreasing
  3. (c) When it is in the direction of velocity
  4. (d) When the object is at rest
Answer & solution

Answer: (c) When it is in the direction of velocity

Correct Answer: When it is in the direction of velocity — Option (c)
Acceleration is a vector quantity (has both magnitude and direction). Acceleration is considered positive when it acts in the same direction as the velocity (or in the positive direction of motion). This means the object is speeding up — its speed is increasing. For example, if a car moves in the positive x-direction and its engine accelerates it forward, the acceleration is positive. Mathematically, if the velocity vector and acceleration vector point in the same direction, the acceleration is positive. This causes the speed to increase over time (positive rate of change of speed).
Option (a) When it is opposite to velocity — Incorrect: When acceleration is opposite to velocity (anti-parallel), it is called negative acceleration or retardation/deceleration; the object slows down.
Option (b) When velocity is decreasing — Incorrect: Decreasing velocity means deceleration/retardation, which corresponds to negative acceleration (acceleration opposing the direction of motion).
Option (d) When the object is at rest — Incorrect: An object at rest has zero velocity; it may have positive, negative, or zero acceleration depending on the net force acting on it. Being at rest alone does not define positive acceleration.
SSC GD 2026 – 27 April 2026 Shift 2 · Numericals on motion

Q59. If a spacecraft travels at 30 km/s, how long will it take to cover 1 AU (150 million km)?

  1. (a) 15 days
  2. (b) 58 days
  3. (c) 100 days
  4. (d) 365 days
Answer & solution

Answer: (b) 58 days

Correct Answer: 58 days — Option (b)
Calculation: Distance = 1 AU = 150,000,000 km; Speed = 30 km/s
Time = Distance / Speed = 150,000,000 km ÷ 30 km/s = 5,000,000 seconds
Converting to days: 5,000,000 ÷ 86,400 (seconds per day) ≈ 57.87 days ≈ 58 days
1 AU (Astronomical Unit) = approximately 150 million km, which is the average distance between the Earth and the Sun. For context, the New Horizons spacecraft (fastest man-made object at the time) travelled at about 16.26 km/s. The Parker Solar Probe has reached speeds of over 160 km/s at perihelion. At 30 km/s, it would take approximately 58 days to travel 1 AU.
Option (a) 15 days — Incorrect: At 30 km/s, the spacecraft would need approximately 58 days to cover 150 million km; 15 days would require a speed of about 115.7 km/s.
Option (c) 100 days — Incorrect: At 30 km/s, the correct time is approximately 58 days; 100 days would imply a speed of only ~17.4 km/s.
Option (d) 365 days — Incorrect: At 30 km/s, a spacecraft would cover 1 AU in about 58 days, not 365 days; 365 days would imply a speed of only ~4.76 km/s.
RRB NTPC GL CBT-1 2025-26 – 18 March 2026 Shift 2 · Distance and displacement

Q60. Which physical quantity can be zero even when a body is in motion?

  1. (a) Distance
  2. (b) Speed
  3. (c) Displacement
  4. (d) Velocity
Answer & solution

Answer: (c) Displacement

Concept: Motion & Physical Quantities
Distance:Total path length, always ≥ 0, never zero in motion.
Speed:Rate of distance covered, cannot be zero if body moves.
Displacement:Shortest distance from initial to final position.
Can be zero if body returns to starting point (circular motion).
Velocity:Rate of displacement, zero only when displacement = 0.
Example:A car completes one full circle → Distance ≠ 0, Displacement = 0.
Answer: Option 3 → Displacement
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q61. A distance–time graph for uniform motion must pass through ________.

  1. (a) The origin
  2. (b) Any point
  3. (c) (1, 1) only
  4. (d) (0, 1)
Answer & solution

Answer: (a) The origin

For uniform motion, an object starts from zero distance at zero time, which means the graph passes through the origin (0,0).
The distance-time graph must begin at the origin to show that at time zero, distance travelled is zero.
Hence, the correct point the graph passes through is the origin.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Distance and displacement

Q62. A farmer walks along the boundary of a square field with side 11 m, completing one round in 20 seconds. After walking for 3 minutes and 20 seconds, what is the magnitude of the displacement from his starting point?

  1. (a) 22 m
  2. (b) 11 m
  3. (c) 15.56 m
  4. (d) 0 m
Answer & solution

Answer: (d) 0 m

The farmer walks along a square field with side 11 m.
One complete round means walking the perimeter of the square.
Perimeter = 4 × side = 4 × 11 = 44 m.
(He completes one round in 20 seconds, so his speed is 44 m)/(20 s) = (2.2 m)/(s.)
Total walking time = 3 minutes 20 seconds = 200 seconds.
Number of complete rounds in 200 seconds = (200)/(20) = 10 rounds.
After 10 complete rounds, he returns to the starting point.
Displacement is the straight-line distance from the starting point to the ending point.
Since he ends at the starting point, displacement = 0 m.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q63. If the velocity-time graph is curved (non-linear), the object is moving with _________.

  1. (a) Uniform acceleration
  2. (b) Constant speed
  3. (c) Constant rate of change of momentum
  4. (d) Non-uniform acceleration
Answer & solution

Answer: (d) Non-uniform acceleration

The velocity-time graph shows how velocity changes over time.
If the graph is curved and not a straight line, acceleration is not constant.
This means the object has non-uniform acceleration.
Uniform acceleration would result in a straight line velocity-time graph.
Constant speed would be a flat, horizontal line on the graph.
Therefore, a curved velocity-time graph implies non-uniform acceleration.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 2 · Numericals on motion

Q64. Reshma swims back and forth in a 120 m long pool, covering a total distance of 240 m in 1 minute. What is her average velocity?

  1. (a) 5 m/s
  2. (b) 0 m/s
  3. (c) 4 m/s
  4. (d) 2 m/s
Answer & solution

Answer: (b) 0 m/s

Reshma swims back and forth in a 120 m pool, total distance = 240 m.
Time taken = 1 minute = 60 seconds.
Displacement = Final position - Initial position = 0 m (since she ends where she started).
Average velocity = (Displacement)/(Time) = (0)/(60) = 0m/s
Average velocity depends on displacement, not total distance.
Hence, the average velocity is 0 m/s
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Distance and displacement

Q65. The location of an object is always described by specifying a point of reference known as the __________.

  1. (a) Origin
  2. (b) Final position
  3. (c) Axis
  4. (d) Displacement
Answer & solution

Answer: (a) Origin

The position of an object is described relative to a fixed point, called the origin or point of reference.
This origin allows us to measure where the object is located in space.
Hence, the correct term for the point of reference is 'Origin'.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Uniform circular motion

Q66. The motion of an athlete running along a circular path with constant speed is an example of accelerated motion because ___________.

  1. (a) The athlete's speed is increasing
  2. (b) The direction of motion changes continuously
  3. (c) A constant external force is applied
  4. (d) The mass of the athlete is constant
Answer & solution

Answer: (b) The direction of motion changes continuously

The athlete runs with constant speed but along a circular path.
In circular motion, velocity direction continuously changes even if speed is constant.
Acceleration depends on change in velocity (magnitude or direction).
Since direction changes, velocity changes, causing acceleration.
Hence, motion is accelerated due to continuous change in direction.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 3 · Motion graphs (distance-time, velocity-time)

Q67. Two bodies A and B are travelling in uniform motion, the slope of the distance- time graph of body A is greater than the slope of the distance - time graph of body B. Which of the following statements is correct?

  1. (a) The speed of body A is less than the speed of body B.
  2. (b) The acceleration of body A is greater than the acceleration of body B.
  3. (c) The speed of body A is equal to the speed of body B.
  4. (d) The speed of body A is greater than speed of body B.
Answer & solution

Answer: (d) The speed of body A is greater than speed of body B.

The slope of a distance-time graph represents the speed of the body.
If the slope of body A's graph is greater than that of body B, it means body A covers more distance in the same time.
Hence, the speed of body A is greater than the speed of body B.
Acceleration is not considered here as both bodies are in uniform motion, implying zero acceleration.
Therefore, the correct statement is: Speed of body A is greater than speed of body B.
RRB Technician Grade III 2025-26 – 10 March 2026 Shift 1 · Uniform circular motion

Q68. Which of the following is the best example of uniform circular motion?

  1. (a) A racing car on a curved track slowing down
  2. (b) A yo-yo being swung in a vertical circle at changing speed
  3. (c) A roller coaster on a loop
  4. (d) Moon revolving around the Earth in a circular orbit
Answer & solution

Answer: (d) Moon revolving around the Earth in a circular orbit

Uniform circular motion means motion in a circle at a constant speed.
Option 1: The car is slowing down, so speed changes, not uniform.
Option 2: Yo-yo speed changes; not uniform.
Option 3: Roller coaster speed varies on loop; not uniform.
Option 4: The Moon moves in nearly circular orbit at almost constant speed; best example.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on motion

Q69. A body covers a straight-line distance of 10 m from point A to B in 5 s and then covers back the distance from point B to point A along the same straight path in 5 s. How much will be the average speed of the body?

  1. (a) 200 m/s
  2. (b) 0.5 m/s
  3. (c) 2 m/s
  4. (d) 5 m/s
Answer & solution

Answer: (c) 2 m/s

The total distance covered is from A to B and back from B to A
10 m + 10 m = 20 m.
The total time taken is 5 s + 5 s = 10 s.
Average speed = (Total distance)/(Total time) = (20 m)/(10 s) = (2 m)/(s.)
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Numericals on motion

Q70. A body moves 6 km towards the east in 1 hour and then 8 km towards the west in the next 1 hour. The average velocity of the body for the whole journey is ___________ .

  1. (a) 1 km/h west
  2. (b) 1 km/h east
  3. (c) 0
  4. (d) 7 km/h east
Answer & solution

Answer: (a) 1 km/h west

Solution:
First 1 hour: 6 km east; next 1 hour: 8 km west.
Net displacement = 6 − 8 =−2 km⇒2 km west.
Total time =2 hours⇒ Average velocity = displacement / time = (−2) / 2 =−1 km/h.
Negative sign shows west direction ⇒1 km/h west.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Numericals on motion

Q71. A body travelling in a straight line with a constant velocity of 50 m/s. What is the displacement of the car in 10 sec and the type of motion followed by the car, respectively?

  1. (a) 500 m and Non-Uniform Motion
  2. (b) 500 m and Uniform Motion
  3. (c) 5 m and Non-Uniform Motion
  4. (d) 5 m and Uniform Motion
Answer & solution

Answer: (b) 500 m and Uniform Motion

(The velocity of the body is given as 50 m)/(s.)
Time duration is given as 10 seconds.
Displacement = velocity × time
= (50 m)/(s × 10 s)
= 500 m
Since velocity is constant, the motion is uniform.
Therefore, the displacement is 500 metres and the motion is uniform.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Numericals on motion

Q72. A bus moves from one end A to the other end B of a straight 1000 m road in 100 s and then turns around and moves 500 m back to point C in another 100 s. What will be the average velocity (in m/s) and the type of motion of the bus in going from point A to point C?

  1. (a) 0.4 m/s and Non-Uniform Motion
  2. (b) 2.5 m/s and Uniform Motion
  3. (c) 2.5 m/s and Non-Uniform Motion
  4. (d) 0.4 m/s and Uniform Motion
Answer & solution

Answer: (c) 2.5 m/s and Non-Uniform Motion

Solution:
Displacement from A to C = 1000 − 500 =500 m
Total time = 100 + 100 =200 s
Average velocity = displacement / time = 500 / 200 =2.5 m/s
Speed changes (goes forward then backward) ⇒Non-uniform motion
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Numericals on motion

Q73. A train starts from rest and reaches a speed of 216 km/h in 10 minutes. If its acceleration is uniform, what is the value of the acceleration?

  1. (a) 0.1 m/s2
  2. (b) 0.6 m/s2
  3. (c) 0.4 m/s2
  4. (d) 1.0 m/s2
Answer & solution

Answer: (a) 0.1 m/s2

Given:
Final speed = 216 km/h
Time = 10 minutes
Acceleration uniform
Step 1: Convert speed to m/s
216 × 1000 / 3600 = 60 m/s
Step 2: Convert time to seconds
10 minutes = 600 s
Step 3: Acceleration formula
a = (v − u) / t
= (60 − 0) / 600
= 0.1 m/s²
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Speed and velocity

Q74. An object moving at 25 m/s is brought to rest in 5 seconds. The final velocity is _________.

  1. (a) 25 m/s
  2. (b) 0 m/s
  3. (c) 20 m/s
  4. (d) 5 m/s
Answer & solution

Answer: (b) 0 m/s

Initial velocity, u = 25 m/s
The object is brought to rest, so final velocity, v = 0 m/s
Rest means no movement, so velocity is zero
Therefore, the final velocity is 0 m/s
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 3 · Numericals on motion

Q75. An object travels 16 m in 4 s and then another 16 m in 2 s. What is the average speed of the object?

  1. (a) 8 m/s
  2. (b) 2.67 m/s
  3. (c) 5.33 m/s
  4. (d) 4 m/s
Answer & solution

Answer: (c) 5.33 m/s

Total distance travelled = 16 m + 16 m = 32 m
Total time taken = 4 s + 2 s = 6 s
Average speed = (Total distance)/(Total time)
Average speed = (32 m)/(6 s) = (5.33 m)/(s)
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 1 · Speed and velocity

Q76. For a body moving with uniform acceleration, average velocity equals ______.

  1. (a) Zero
  2. (b) (Initial velocity + Final velocity)/2
  3. (c) Initial velocity
  4. (d) Final velocity
Answer & solution

Answer: (b) (Initial velocity + Final velocity)/2

Average velocity under uniform acceleration is the mean of initial and final velocities.
This means average velocity = (Initial velocity + Final velocity) / 2.
Hence, option 2 is correct.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q77. If a velocity–time graph shows a horizontal straight line, the object is moving with ________.

  1. (a) Non-uniform acceleration
  2. (b) Constant velocity
  3. (c) Zero displacement
  4. (d) No velocity
Answer & solution

Answer: (b) Constant velocity

A horizontal straight line on a velocity-time graph means velocity is constant over time.
Constant velocity implies zero acceleration.
Thus, the object moves with constant velocity when the velocity-time graph is a horizontal line.
RRB Technician Grade III 2025-26 – 9 March 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q78. In a velocity–time graph, the graph is a straight horizontal line at 10 m/s for 5 seconds. Then the distance travelled is ____________.

  1. (a) 10 m
  2. (b) 5 m
  3. (c) 50 m
  4. (d) 2 m
Answer & solution

Answer: (c) 50 m

(Velocity is constant at 10 m)/(s for 5 seconds.)
Distance = Velocity × Time
Distance = (10 m)/(s × 5 s)
Distance = 50 m
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Uniform circular motion

Q79. A cyclist riding at constant speed on a circular track is an example of which type of motion?

  1. (a) Non-uniform motion
  2. (b) Uniform circular motion
  3. (c) Linear motion
  4. (d) Oscillatory motion
Answer & solution

Answer: (b) Uniform circular motion

A cyclist riding at constant speed on a circular track moves in a circle with uniform speed.
This is an example of uniform circular motion.
The speed is constant but the direction changes continuously.
Hence, the motion is uniform circular motion.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Motion graphs (distance-time, velocity-time)

Q80. For an object moving with uniform velocity, the height of its velocity-time graph remains constant with time. This results in a graph that is a _________.

  1. (a) Straight line parallel to the x-axis
  2. (b) Curve
  3. (c) Vertical line
  4. (d) Straight line passing through the origin
Answer & solution

Answer: (a) Straight line parallel to the x-axis

For an object moving with uniform velocity, velocity does not change with time.
Therefore, velocity-time graph is a straight horizontal line.
A straight line parallel to the x-axis shows constant velocity over time.
So, the velocity-time graph is a straight line parallel to the x-axis.
Final answer: Option 1
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Equations of motion

Q81. When applying the equations of motion to an object moving under gravitational force and thrown vertically upwards, the acceleration (a) should be taken as ______.

  1. (a) Positive, as it is in the direction of velocity.
  2. (b) Negative, as it opposes the initial motion.
  3. (c) Positive, as gravity is constant.
  4. (d) Independent of the direction of motion.
Answer & solution

Answer: (b) Negative, as it opposes the initial motion.

The acceleration due to gravity always acts downwards, towards the Earth’s centre.
When an object is thrown vertically upwards, its initial velocity is upwards, but acceleration due to gravity acts in the opposite direction, downwards.
Hence, acceleration is negative with respect to the upward direction of velocity.
Therefore, acceleration (a) should be taken as negative since it opposes the initial motion upwards.
Final answer: Option 2
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 3 · Speed and velocity

Q82. Which characteristic defines non-uniform motion?

  1. (a) Equal displacement in equal time intervals
  2. (b) Constant speed
  3. (c) Covering unequal distances in equal intervals of time
  4. (d) Zero acceleration
Answer & solution

Answer: (c) Covering unequal distances in equal intervals of time

Non-uniform motion means the object does not move equal distances in equal time intervals.
It covers unequal distances in equal intervals of time.
This is because speed or velocity is changing.
So, non-uniform motion is defined by covering unequal distances in equal time intervals.
Therefore, the correct answer is option 3.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 2 · Motion graphs (distance-time, velocity-time)

Q83. Which of the following is/are NOT true about the distance-time graph for a body moving in uniform motion?
(i) The slope of the distance-time graph gives the acceleration of the body.
(ii) The lesser the slope of the graph the slower the body is moving.
(iii) The area under this graph gives the speed of the body.

  1. (a) Both (i) and (iii)
  2. (b) Only (iii)
  3. (c) Only (ii)
  4. (d) Both (i) and (ii)
Answer & solution

Answer: (a) Both (i) and (iii)

Option 1 is correct.
(i) The slope of the distance-time graph gives the speed, not acceleration. Acceleration is the slope of velocity-time graph.
(ii) The lesser slope indicates slower speed, so this is true.
(iii) The area under the distance-time graph does not represent speed; speed is given by the slope.
Hence, both (i) and (iii) are NOT true about distance-time graph for uniform motion.
Answer ends here.
RRB Technician Grade III 2025-26 – 6 March 2026 Shift 1 · Acceleration and retardation

Q84. Which of the following situations shows zero acceleration?

  1. (a) An aeroplane taking off
  2. (b) A car moving with constant speed on a straight road
  3. (c) A train speeding up
  4. (d) A ball thrown upward
Answer & solution

Answer: (b) A car moving with constant speed on a straight road

Option 2 is correct: A car moving with constant speed on a straight road shows zero acceleration.
Acceleration means change in velocity.
If the speed and direction are constant, acceleration is zero.
An aeroplane taking off, a train speeding up, and a ball thrown upward all experience a change in velocity, hence non-zero acceleration.
Therefore, only the car moving at constant speed in a straight line has zero acceleration.
Final answer: Option 2
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Numericals on motion

Q85. A cyclist increases her speed from 6 m/s to 12 m/s in 3 seconds on a straight road. What is her acceleration during this period?

  1. (a) 3 m/s²
  2. (b) 2 m/s²
  3. (c) 6 m/s²
  4. (d) 0.5 m/s²
Answer & solution

Answer: (b) 2 m/s²

Initial speed (u) = 6 m/s
Final speed (v) = 12 m/s
Time taken (t) = 3 seconds
Acceleration (a) =(v - u)/(t)
a =(12 - 6)/(3)
a =(6)/(3) = 2 m/s²
Re-check calculation: 6 to 12 in 3 seconds means speed doubles in 3 seconds; acceleration is (final speed − initial speed)/time
a =(12 - 6)/(3)= 2 m/s²
Correct option is 2 m/s²
Final answer: Option 1
RRB JE CBT-1 2025-26 – 20 February 2026 Shift 2 · Speed and velocity

Q86. Which of the following does NOT define the motion of a pendulum swinging back and forth?

  1. (a) The pendulum exhibits uniform motion
  2. (b) The pendulum exhibits periodic motion
  3. (c) The pendulum exhibits non-uniform motion
  4. (d) The pendulum exhibits accelerated motion
Answer & solution

Answer: (a) The pendulum exhibits uniform motion

The motion of a pendulum is not uniform because its speed varies during the swinging.
A pendulum swinging back and forth exhibits periodic motion, meaning it repeats at regular intervals.
The motion is non-uniform due to changing speed, so it exhibits non-uniform motion.
It also exhibits accelerated motion as acceleration changes direction during the swing.
Hence, the pendulum does NOT exhibit uniform motion.
Final answer: Option 1
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 3 · Speed and velocity

Q87. A car travels 120 meters north in 10 seconds and then instantly returns to its original position in the next 10 seconds. Which statement best describes the car's average velocity for the entire trip?

  1. (a) The average velocity is 12 m/s because the car covered 120 m in 10 seconds.
  2. (b) The average velocity is 12 m/s because the total time is 10 seconds.
  3. (c) The average velocity is 6 m/s because the total distance is 240 m in 20 seconds.
  4. (d) The average velocity is zero because the displacement is zero.
Answer & solution

Answer: (d) The average velocity is zero because the displacement is zero.

The car goes120 m northand thenreturns 120 m southto the starting point.
So thetotal displacement = 0 m.
Average velocity =(Total displacement)/(Total time).
Since displacement iszero, theaverage velocity = 0 m/s.
Answer: 0 m/s
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Numericals on motion

Q88. If two cars start at the same point and move along a straight road, Car X at a constant speed of 40 km/h and Car Y at a constant speed of 60 km/h, which statement correctly compares their motions after 2 hours?

  1. (a) Car Y will be 20 km ahead of Car X after 2 hours.
  2. (b) Both cars will have traveled the same distance after 2 hours.
  3. (c) Car X will be 40 km ahead of Car Y after 2 hours.
  4. (d) Car Y will be 40 km ahead of Car X after 2 hours.
Answer & solution

Answer: (d) Car Y will be 40 km ahead of Car X after 2 hours.

Solution:
Distance = Speed × Time
Car X:
40 × 2 =80 km
Car Y:
60 × 2 =120 km
Difference =
120 − 80 =40 km
SoCar Y is 40 km ahead of Car X.
Answer: Car Y will be 40 km ahead of Car X after 2 hours.
RRB JE CBT-1 2025-26 – 19 February 2026 Shift 2 · Uniform circular motion

Q89. In the context of uniform circular motion, which of the following examples from everyday life best illustrates this concept?

  1. (a) A car moving with changing speed along a circular track
  2. (b) A person walking randomly in a park
  3. (c) A satellite revolving around the Earth at a fixed speed in a circular orbit
  4. (d) A pendulum bob swinging to and fro
Answer & solution

Answer: (c) A satellite revolving around the Earth at a fixed speed in a circular orbit

The concept of uniform circular motion refers to an object moving in a circle at a constant speed.
Option 1 is incorrect because the car's speed is changing, so the motion is not uniform.
Option 2 describes random walking, not circular motion.
Option 3 correctly describes a satellite moving at a constant speed along a circular orbit around the Earth, which is uniform circular motion.
Option 4 describes a pendulum, which moves in a back-and-forth arc, not a full circular path at constant speed.
Hence, option 3 best illustrates uniform circular motion.
Final answer: Option 3
SSC MTS 2025 – 18 February 2026 Shift 1 · Distance and displacement

Q90. If the final position of an object coincides with the initial position of the object then, the displacement of the object is _____.

  1. (a) zero
  2. (b) double
  3. (c) half
  4. (d) one-fourth
Answer & solution

Answer: (a) zero

Displacement is defined as the change in position of an object: Displacement = Final Position − Initial Position. If the final position coincides with the initial position, then Displacement = 0 (zero). This is the case, for example, when an object completes a full circular path. Double, half, and one-fourth are all incorrect.
SSC MTS 2025 – 9 February 2026 Shift 3 · Uniform circular motion

Q91. What of the following is an example of object moving under uniform circular motion?
I. Motion of the Moon
II. Motion of the Earth

  1. (a) Only I
  2. (b) Only II
  3. (c) Both I and II
  4. (d) Neither I nor II
Answer & solution

Answer: (c) Both I and II

Correct Answer: 3 – Both I and II.
Uniform circular motion refers to motion in a circular path at a constant speed. Both the Moon (orbiting the Earth) and the Earth (orbiting the Sun) approximately follow circular orbits at nearly constant speeds and are standard examples of uniform circular motion used in physics.
Option 1 (Only I) is incorrect as Earth's motion is also an example.
Option 2 (Only II) is incorrect as Moon's motion is equally valid.
Option 4 (Neither I nor II) is incorrect as both are classic examples of uniform circular motion.
Delhi Police Head Constable (Ministerial) 2025-26 – 10 January 2026 Shift 2 · Projectile and relative motion

Q92. While sitting on a merry-go-round turning anti-clockwise, surrounding objects appear to move in which direction?

  1. (a) Clockwise
  2. (b) Anti-clockwise
  3. (c) Random
  4. (d) Not moving
Answer & solution

Answer: (a) Clockwise

Clockwise is correct. This is due to the principle of relative motion. When you are sitting on a merry-go-round rotating anti-clockwise, your frame of reference is rotating anti-clockwise. Stationary surrounding objects, when observed from this rotating frame, appear to move in the opposite direction — i.e., clockwise.
• Anti-clockwise: Objects would appear anti-clockwise only if the merry-go-round itself were rotating clockwise.
• Random: Motion is not random; it follows a definite pattern based on the direction of rotation.
• Not moving: Objects would appear stationary only if the observer were also stationary (non-rotating).
SSC CHSL Tier-1 2025 – 22 November 2025 Shift 1 · Speed and velocity

Q93. The object covers unequal distances in equal intervals of time, it is said to be in _____ motion.

  1. (a) uniform
  2. (b) linear
  3. (c) non-uniform
  4. (d) circular
Answer & solution

Answer: (c) non-uniform

Correct Answer: non-uniform (Option 3)
When an object covers unequal distances in equal intervals of time, it is said to be in non-uniform motion (also called accelerated or decelerated motion). This means the speed of the object is changing — it is either speeding up or slowing down. Examples include a car braking, a ball rolling down a slope, or a ball thrown upward.
Why other options are wrong:
Option 1 - uniform motion: In uniform motion, an object covers EQUAL distances in equal intervals of time (constant speed).
Option 2 - linear motion: Linear motion refers to movement in a straight line and does not specify whether the speed is constant or changing.
Option 4 - circular motion: Circular motion refers to movement along a circular path, not to the equality of distances covered.
SSC CGL Tier-1 2025 – 21 September 2025 Shift 2 · Rotational motion and moment of inertia

Q94. The moment of inertia of a rigid body about an axis:

  1. (a) (a) Is a scalar quantity, independent of the axis
  2. (b) (b) Is a vector quantity, dependent on the axis
  3. (c) (c) Is a scalar quantity, dependent on the axis
  4. (d) (d) Is a tensor quantity, independent of the axis
Answer & solution

Answer: (c) (c) Is a scalar quantity, dependent on the axis

Moment of inertia (I) = Σmr², a scalar quantity.
It depends on the distribution of mass relative to the chosen axis of rotation.
Hence it is a scalar quantity that is dependent on the axis.
SSC CGL Tier-1 2025 – 12 September 2025 Shift 1 · Rotational motion and moment of inertia

Q95. For a rigid body, the angular velocity of any particle about a given axis of rotation is:

  1. (a) Proportional to its distance from the axis
  2. (b) The same for all particles
  3. (c) Inversely proportional to its distance from the axis
  4. (d) Always zero
Answer & solution

Answer: (b) The same for all particles

In a rigid body rotating about a fixed axis, every particle has the same angular velocity (ω).
Linear velocity v = rω varies with distance, but ω is identical for every particle.
Hence, the answer is option 2.
SSC Selection Post 2025 (Phase XIII) – 31 July 2025 Shift 3 · Numericals on motion

Q96. A car accelerates uniformly from rest to a speed of 20 m/s in 5 seconds. What is the magnitude of its acceleration?

  1. (a) 2 m/s²
  2. (b) 4 m/s²
  3. (c) 10 m/s²
  4. (d) 25 m/s²
Answer & solution

Answer: (b) 4 m/s²

Correct Answer: Option 2 – 4 m/s²
Using the formula for uniform acceleration: a = (v − u) / t
Where: v = final velocity = 20 m/s; u = initial velocity = 0 m/s (starts from rest); t = time = 5 seconds
a = (20 − 0) / 5 = 20 / 5 = 4 m/s²
Option 1 (2 m/s²): Would require 10 seconds to reach 20 m/s — not the correct calculation.
Option 3 (10 m/s²): Would require only 2 seconds to reach 20 m/s — not the correct calculation.
Option 4 (25 m/s²): This would be the final speed (25 m/s) confused with acceleration, or an incorrect formula application.
SSC Selection Post 2025 (Phase XIII) – 28 July 2025 Shift 2 · Acceleration and retardation

Q97. Which of the following statements best describes the relationship between velocity and acceleration in different dimensions?

  1. (a) In one-dimensional motion, velocity and acceleration are always perpendicular to each other.
  2. (b) In two-dimensional motion, acceleration must always act opposite to velocity.
  3. (c) In one-dimensional motion, velocity and acceleration lie along the same line, either in the same or opposite direction.
  4. (d) In three-dimensional motion, velocity and acceleration must always act in the same direction.
Answer & solution

Answer: (c) In one-dimensional motion, velocity and acceleration lie along the same line, either in the same or opposite direction.

Correct Answer: Option 3 – In one-dimensional motion, velocity and acceleration lie along the same line, either in the same or opposite direction
In one-dimensional (1D) motion, all quantities are confined to a single straight line. Therefore, velocity and acceleration must lie along this same line — they are either parallel (same direction, causing speeding up) or anti-parallel (opposite directions, causing slowing down). They cannot be perpendicular in 1D. This is a fundamental constraint of one-dimensional kinematics.
Option 1 (Perpendicular in 1D): Impossible in 1D — there is only one direction. Perpendicular vectors require at least 2D.
Option 2 (Acceleration always opposite to velocity in 2D): False — in 2D, acceleration can be in any direction relative to velocity (e.g., circular motion has centripetal acceleration perpendicular to velocity).
Option 4 (Must act in same direction in 3D): False — in 3D, acceleration can be in any direction relative to velocity (e.g., projectile motion, circular motion).
SSC Selection Post 2025 (Phase XIII) – 25 July 2025 Shift 2 · Uniform circular motion

Q98. A stone is moving in a circular path with constant speed. Which of the following statements about its motion is correct?

  1. (a) The stone has zero acceleration because its speed is constant.
  2. (b) The direction of the stone's velocity remains unchanged.
  3. (c) The stone experiences a constant acceleration directed towards the centre of the circle
  4. (d) The motion of the stone is considered non-uniform because its speed varies.
Answer & solution

Answer: (c) The stone experiences a constant acceleration directed towards the centre of the circle

Correct Answer: Option 3 – The stone experiences a constant acceleration directed towards the centre of the circle
In uniform circular motion, the speed is constant but the direction of velocity continuously changes. Since velocity (a vector) changes direction, there IS acceleration — called centripetal acceleration — directed towards the centre of the circular path. The magnitude of centripetal acceleration = v²/r (constant, since v and r are constant), but its direction continuously changes as the stone moves around the circle.
Option 1 (Zero acceleration): Incorrect. Although speed is constant, the change in direction of velocity means acceleration exists (centripetal acceleration).
Option 2 (Direction unchanged): Incorrect. In circular motion, the direction of velocity (tangent to the circle) continuously changes at every point.
Option 4 (Non-uniform speed): Incorrect. In uniform circular motion, speed is constant (only direction changes); the motion is uniform in terms of speed.
SSC Selection Post 2025 (Phase XIII) – 24 July 2025 Shift 1 · Distance and displacement

Q99. What is the difference between displacement and distance?

  1. (a) Distance includes direction; displacement does not
  2. (b) Displacement includes direction; distance does not
  3. (c) Both are scalar quantities
  4. (d) Both are vector quantities
Answer & solution

Answer: (b) Displacement includes direction; distance does not

Correct Answer: Option 2 – Displacement includes direction; distance does not
Displacement is a vector quantity — it has both magnitude and direction (shortest straight-line path from start to end point). Distance is a scalar quantity — it has only magnitude (total path length travelled) with no direction. For example, if you walk 4 km north then 3 km east, distance = 7 km but displacement = 5 km (northeast).
Option 1 (Distance includes direction): Incorrect; distance is scalar with no direction.
Option 3 (Both scalar): Incorrect; displacement is a vector, not scalar.
Option 4 (Both vector): Incorrect; distance is scalar, not a vector.
RRB Technician Grade III 2024-25 – 29 December 2024 Shift 3 · Distance and displacement

Q100. The shortest distance measured from the initial to the final position of an object is known as _______________.

  1. (a) acceleration
  2. (b) velocity
  3. (c) displacement
  4. (d) speed
Answer & solution

Answer: (c) displacement

Correct Answer: Option 3 — displacement
Displacement is defined as the shortest distance between the initial and the final position of a moving object, measured in a definite direction. It is therefore a vector quantity: it has both magnitude and direction. Its SI unit is the metre (m).
Displacement must be carefully distinguished from distance. Distance is the total length of the actual path covered, is a scalar, and can never be negative or zero for a moving body. Displacement, on the other hand, can be positive, negative or even zero. If a boy runs once around a circular track and comes back to the starting point, his distance equals the circumference of the track, but his displacement is zero.
Also, displacement ≤ distance always. The two are equal only when the body moves along a straight line in one direction without turning back.
❌ Why the other options are incorrect
✗ acceleration — the rate of change of velocity, a = (v − u)/t, measured in m s⁻², not a length at all.
✗ velocity — displacement per unit time, measured in m/s; it is a rate, whereas displacement is a length.
✗ speed — distance per unit time, a scalar measured in m/s; again a rate, not the shortest length between two positions.
📌 SSC & Railway Exam Facts
• Distance — scalar, total path length; Displacement — vector, shortest straight-line length between the two positions.
• |Displacement| ≤ Distance; they are equal only for motion in a straight line without reversal.
• Speed = distance/time (scalar); Velocity = displacement/time (vector); both in m/s.
• For one complete circular round, displacement = 0 but distance = 2πr.
• In a displacement–time graph, the slope gives velocity; in a velocity–time graph, the slope gives acceleration and the area gives displacement.
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