Motion PYQs – Physics, SSC & Railway

Correct Answer: Option 3 — uniform circular motion
When a body moves along a circular path covering equal distances in equal intervals of time, its speed stays constant and the motion is called uniform circular motion.
The important point for exams is that even though the speed is uniform, the motion is accelerated. Velocity is a vector, and at every point of the circle the direction of motion (along the tangent) keeps changing. A changing direction means a changing velocity, and therefore an acceleration.
This acceleration is directed towards the centre of the circle and is called centripetal acceleration, a = v²/r. The force that produces it is the centripetal force, F = mv²/r, supplied by tension in a string, friction on a road, or gravitation for a satellite. If this force suddenly disappears — say the string breaks — the body flies off along the tangent in a straight line, exactly as Newton's first law demands.
The speed in uniform circular motion is v = 2πr/t, where t is the time for one complete revolution.
❌ Why the other options are incorrect
✗ linear motion — motion along a line, in which the direction does not change; here the direction changes continuously.
✗ motion along a straight line — the same idea again; a circular path is by definition not a straight line.
✗ motion along a hemisphere — not a standard term in kinematics, and the question clearly describes a circular path.
📌 SSC & Railway Exam Facts
• Uniform circular motion — speed constant, velocity changing, hence an accelerated motion.
• v = 2πr/t; centripetal acceleration a = v²/r; centripetal force F = mv²/r, directed towards the centre.
• Centripetal force does zero work, because it is always perpendicular to the displacement.
• Examples — a satellite around the Earth, an athlete on a circular track, a stone whirled on a string, the blades of a fan.
• When the string breaks, the stone moves along the tangent — an application of inertia of direction.

Correct Answer: Option 4 — 0m, 10m
📖 Given
• Maximum height reached upward, h = 5 m
• The body then falls back to the point from which it was thrown.
✍️ Formula
Distance = total length of the path actually travelled (a scalar)
Displacement = shortest straight line from the initial to the final position (a vector)
🔢 Solution
• Path travelled going up = 5 m
• Path travelled coming down = 5 m
• Total distance = 5 + 5 = 10 m
• The body ends exactly where it started, so initial and final positions are the same.
• Total displacement = 0 m
The body has certainly moved 10 m of path, but since it returned to the starting point, its net change of position is zero.
❌ Why the other options are incorrect
✗ 5m, 5m — this is the answer only for the upward journey alone; the return trip has been ignored.
✗ 0m, 0m — the displacement is right, but the distance can never be zero when a body has actually moved along a path.
✗ 10m, 0m — the two values have been interchanged; 10 m is the distance and 0 m is the displacement.
📌 SSC & Railway Exam Facts
• Distance is a scalar and is always positive; displacement is a vector and can be positive, negative or zero.
• Distance ≥ Displacement always; they are equal only for motion in a straight line without turning back.
• For one complete round of a circle of radius r: distance = 2πr, displacement = 0.
• SI unit of both distance and displacement is the metre (m).
• Average speed = total distance / total time; average velocity = total displacement / total time.

Correct Answer: Option 3 — 0 m
📖 Given
• Distance between A and B = 20 m
• Path followed: A → B → A
• The object finally comes back to its starting point A
✍️ Formula
Displacement = shortest distance from the initial position to the final position (a vector)
🔢 Solution
• Initial position = A
• Final position = A (the object returns)
• Since the initial and final positions are the same point, the straight-line gap between them is 0
• Net displacement = 0 m
Note that the total distance travelled is 20 + 20 = 40 m, but distance and displacement are not the same thing.
❌ Why the other options are incorrect
✗ 10 m — this would be half of AB; the object does not stop midway, it returns to A.
✗ 20 m — this is the displacement only for the one-way trip A → B, not for the full journey.
✗ 40 m — this is the total distance covered (20 m going + 20 m returning), not the displacement.
📌 SSC & Railway Exam Facts
• Distance is a scalar and can never be negative or zero for a moving body; displacement is a vector and can be zero.
• Displacement ≤ distance, always; they are equal only for straight-line motion in one direction.
• For a body completing one full circle of radius r: distance = 2πr, displacement = 0.
• Average speed = total distance/total time; average velocity = net displacement/total time.
• Here the average velocity for the whole trip is also zero.

Correct Answer: Option 1 — 2 m s⁻¹
📖 Given
• First part: distance s₁ = 60 m in time t₁ = 30 s
• Second part: distance s₂ = 80 m in time t₂ = 40 s
✍️ Formula
Average speed = total distance / total time
🔢 Solution
• Total distance = 60 + 80 = 140 m
• Total time = 30 + 40 = 70 s
• Average speed = 140/70
• Average speed = 2 m s⁻¹
So over the whole journey the bus covers 2 metres every second, even though its speed in each separate part was exactly 2 m s⁻¹ as well.
❌ Why the other options are incorrect
✗ 1 m s⁻¹ — half the correct value; you get it by dividing 70 s by 140 m the wrong way, or by using only one part of the trip.
✗ 0.15 m s⁻¹ — this is 70/(60 × 80)-type nonsense; time must be divided into distance, not the other way round.
✗ 2.28 m s⁻¹ — this is 160/70, obtained by wrongly adding the distances as 80 + 80 instead of 60 + 80.
📌 SSC & Railway Exam Facts
• Average speed = total distance/total time; it is not the plain average of the two speeds unless the times are equal.
• Average velocity = net displacement/total time, a vector.
• SI unit of speed is m s⁻¹; 1 km/h = 5/18 m s⁻¹.
• Uniform speed means equal distances in equal intervals of time.
• Slope of a distance–time graph gives the speed.

Correct Answer: Option 4 — πr metre/second
📖 Given
• Radius of the circular path = r
• Time for one complete round, t = 2 s
✍️ Formula
v = distance/time, and the distance in one round is the circumference = 2πr, so v = 2πr/t
🔢 Solution
• v = 2πr/t
• v = 2πr/2
• v = πr metre/second
So an object going once round a circle of radius r in 2 seconds moves with a speed of πr metres every second. Note that although the speed is constant, the velocity keeps changing because the direction of motion changes at every point.
❌ Why the other options are incorrect
✗ πr/2 metre/second — this comes from dividing the circumference by 2 twice, i.e. using (2πr/2)/2.
✗ πr/4 metre/second — this divides the circumference by 8 in effect; no step of the problem gives a factor of 4 in the denominator.
✗ 2πr metre/second — this is the circumference itself, i.e. the distance covered; it would be the speed only if the time taken were 1 second.
📌 SSC & Railway Exam Facts
• For uniform circular motion, v = 2πr/t, where t is the time period.
• Circumference of a circle = 2πr; area = πr².
• In circular motion speed may be constant but velocity is always changing, so the motion is accelerated.
• The force that keeps a body on a circular path is the centripetal force, F = mv²/r, directed towards the centre.
• Work done by the centripetal force in one full circle is zero.

Correct Answer: Option 3 — 100 m
📖 Given
• Retardation (acceleration opposite to motion), a = −2 m s⁻²
• Time taken to stop, t = 10 s
• Final velocity, v = 0 (the cycle stops)
✍️ Formula
First find u from v = u + at, then use s = ut + ½at²
🔢 Solution
• 0 = u + (−2 × 10)
• 0 = u − 20
• u = 20 m/s
• s = (20 × 10) + ½ × (−2) × (10)²
• s = 200 − (1 × 100)
• s = 200 − 100
• s = 100 m
So Ram's cycle rolls another 100 metres after he squeezes the brakes before it finally comes to rest.
❌ Why the other options are incorrect
✗ 150 m — no combination of the given data gives this; it is a made-up value between 100 and 200.
✗ 50 m — this comes from using ½at² = ½ × 2 × 10² ÷ 2 wrongly, i.e. halving the answer once more.
✗ 200 m — this is only the first term ut = 20 × 10; the student forgot to subtract ½at² = 100 m.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• A quicker route here is s = (u + v)/2 × t = (20 + 0)/2 × 10 = 100 m.
• Retardation is negative acceleration — the velocity decreases with time.
• SI unit of acceleration is m/s² (m s⁻²); it is a vector quantity.
• Stopping distance ∝ u², which is why doubling the speed makes braking distance four times longer.

Correct Answer: Option 4 — The object moves with a constant speed.
An object is said to be in uniform motion when it covers equal distances in equal intervals of time, however small those intervals are taken. Covering equal distances in equal times is exactly the same as saying that the speed stays constant.
Because the speed does not change, a body in uniform motion (moving along a straight line) has zero acceleration. Its distance–time graph is a straight line, whose slope gives the speed, and its speed–time graph is a straight line parallel to the time axis. In non-uniform motion, unequal distances are covered in equal times, the speed keeps changing, and the acceleration is not zero.
❌ Why the other options are incorrect
✗ The object travels with different acceleration. — in uniform motion along a straight line the acceleration is zero and constant; a changing acceleration means the motion is definitely non-uniform.
✗ The object moves with varying speeds. — a varying speed is the very definition of non-uniform motion, the opposite of what is asked.
✗ The object does not travel equal distances with respect to equal time intervals. — this again describes non-uniform motion; uniform motion requires equal distances in equal times.
📌 SSC & Railway Exam Facts
• Uniform motion — equal distances in equal time intervals; speed constant, acceleration = 0.
• Distance–time graph: straight line for uniform motion; its slope = speed.
• Velocity–time graph: its slope = acceleration, and the area under it = distance travelled.
• Speed is a scalar, velocity a vector; SI unit m/s; 1 km/h = 5/18 m/s.
• In uniform circular motion the speed is constant but the velocity changes, so the motion is accelerated.

Correct Answer: Option 3 — 630 m
📖 Given
• Initial velocity, u = 12 m s⁻¹
• Final velocity, v = 30 m s⁻¹
• Time taken, t = 30 s
• Acceleration is uniform
✍️ Formula
a = (v − u)/t and s = ut + ½at²
🔢 Solution
• a = (30 − 12)/30
• a = 18/30 = 0.6 m s⁻²
• s = (12 × 30) + ½ × 0.6 × (30)²
• s = 360 + (0.3 × 900)
• s = 360 + 270
• s = 630 m
The same answer comes straight from the average-velocity shortcut: s = ((u + v)/2) × t = ((12 + 30)/2) × 30 = 21 × 30 = 630 m.
❌ Why the other options are incorrect
✗ 175 m — does not follow from any correct step of this problem; it is far too small for a bus moving at 12–30 m s⁻¹ for half a minute.
✗ 400 m — this is roughly what you get if only the change in velocity (18 m s⁻¹) is used with a wrong time factor.
✗ 1900 m — this comes from wrongly multiplying the final velocity by a much larger time, ignoring that the bus started at 12 m s⁻¹ and ran for only 30 s.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• For uniform acceleration, average velocity = (u + v)/2.
• Area under a velocity–time graph gives the distance travelled.
• Slope of a velocity–time graph gives the acceleration.
• SI unit of acceleration is m s⁻²; it is a vector quantity.

Correct Answer: Option 2 — 150 m
📖 Given
• Initial velocity, u = 0 (starts from rest)
• Final velocity, v = 30 m s⁻¹
• Time taken, t = 10 s
• Acceleration is uniform
✍️ Formula
a = (v − u)/t and s = ut + ½at²
🔢 Solution
• a = (30 − 0)/10 = 3 m s⁻²
• s = (0 × 10) + ½ × 3 × (10)²
• s = 0 + (1.5 × 100)
• s = 150 m
The average-velocity shortcut gives the same result: s = ((0 + 30)/2) × 10 = 15 × 10 = 150 m. So the motorcycle covers 150 metres while building up to 30 m s⁻¹.
❌ Why the other options are incorrect
✗ 250 m — not obtained from any correct step; it would need an acceleration of 5 m s⁻² over 10 s.
✗ 270 m — this is the answer to a different problem (u = 12 m s⁻¹ type data); it does not fit u = 0.
✗ 300 m — this is v × t = 30 × 10, which wrongly assumes the motorcycle moved at the full 30 m s⁻¹ for the whole 10 s; in fact it started from rest, so its average speed was only 15 m s⁻¹.
📌 SSC & Railway Exam Facts
• Equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• Starting from rest means u = 0, which simplifies s = ½at².
• For uniform acceleration, average velocity = (u + v)/2.
• Area under the velocity–time graph = distance travelled.
• Distance is a scalar; displacement is a vector.

Correct Answer: Option 4 — 0 m/s
📖 Given
• Length of the pool = 60 m
• Total distance covered = 120 m (60 m one way + 60 m back)
• Total time, t = 1 minute = 60 s
• She returns to her starting point, so displacement = 0
✍️ Formula
Average velocity = Total displacement / Total time
(while Average speed = Total distance / Total time)
🔢 Solution
• Displacement = 60 m (forward) − 60 m (backward) = 0 m
• Average velocity = 0/60
• Average velocity = 0 m/s
• (For comparison, average speed = 120/60 = 2 m/s)
Because Uma finishes exactly where she started, her net change of position is zero, so her average velocity is zero even though she has actually swum 120 m.
❌ Why the other options are incorrect
✗ 2 m/s — this is the average speed (120 m ÷ 60 s); speed uses the distance, velocity uses the displacement.
✗ 1 m/s — this comes from using only the one-way length, 60 m ÷ 60 s, but she does not stop at the far end; she comes back.
✗ 3 m/s — no correct step gives this value; it would need a distance of 180 m in one minute, which is not what the question states.
📌 SSC & Railway Exam Facts
• Distance is a scalar and can never decrease; displacement is a vector and can be zero or negative.
• Average speed = total distance/total time; average velocity = total displacement/total time; SI unit of both is m/s.
• For any closed path (round trip) the displacement, and hence the average velocity, is always zero.
• |Displacement| ≤ distance; the two are equal only for motion in a straight line without turning back.
• 1 km/h = 5/18 m/s and 1 m/s = 18/5 km/h.

Correct Answer: Option 1 — 2πR/t
📖 Given
• Radius of the circular track, R
• Time taken for one complete round, t s
• The speed is uniform
✍️ Formula
Speed = Total distance travelled / Time taken, and for one full circle the distance is the circumference = 2πR
🔢 Solution
• Distance covered in one round = 2πR
• Time taken = t s
• Speed, v = 2πR / t
• v = 2πR/t
In plain language, the athlete runs the whole circumference of the track in t seconds, so his speed is that circumference divided by the time.
❌ Why the other options are incorrect
✗ πR/2t — this is a quarter of the correct value; it appears if the circumference is wrongly taken as πR/2.
✗ 3πR/t — uses a distance of 3πR, which is one-and-a-half circles, not one round.
✗ πR/t — this uses πR, the length of only half the circle (a semicircle), so it gives half the true speed.
📌 SSC & Railway Exam Facts
• Circumference of a circle = 2πr; area = πr².
• In one complete round the distance = 2πR but the displacement = 0, so the average velocity is zero while the average speed is 2πR/t.
• Speed is a scalar, velocity is a vector; SI unit of both is m/s.
• In uniform circular motion the speed is constant but the velocity keeps changing because the direction changes — so the motion is accelerated.
• Centripetal acceleration a = v²/R, directed towards the centre.

Correct Answer: Option 2 — 0.4 m/s²
📖 Given
• Initial velocity, u = 0 (he starts from rest)
• Final velocity, v = 24 m/s
• Time, t = 1 minute = 60 s
✍️ Formula
a = (v − u)/t
🔢 Solution
• a = (24 − 0)/60
• a = 24/60
• a = 0.4 m/s²
This means Raju's speed grows by 0.4 metre per second in every second of pedalling.
❌ Why the other options are incorrect
✗ 0 m/s² — zero acceleration means the velocity never changes; but Raju goes from 0 to 24 m/s, so the acceleration cannot be zero.
✗ 0.6 m/s² — this comes from dividing 24 by 40 instead of by 60 s.
✗ 0.2 m/s² — this comes from taking the time as 2 minutes (120 s) instead of 1 minute.
📌 SSC & Railway Exam Facts
• Acceleration is the rate of change of velocity; SI unit m/s², and it is a vector quantity.
• The three equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• Always convert first — 1 minute = 60 s, 1 hour = 3600 s.
• When the final velocity is less than the initial velocity, the acceleration is negative and is called retardation.
• 1 km/h = 5/18 m/s and 1 m/s = 18/5 km/h.

Correct Answer: Option 4 — Only (C)
The equation of motion that links final velocity v, initial velocity u, acceleration a and time t is the first equation of motion:
v = u + at
Taking u to the left-hand side gives (v − u) = at, which is exactly statement (C). For a freely falling body the acceleration is g, so it becomes v = u + gt. Hence only (C) is correct.
❌ Why the other options are incorrect
✗ Statement (A) — s = ut + 2at² — the correct second equation is s = ut + ½at²; the coefficient must be ½, not 2, and in any case it gives distance, not a velocity–time relation.
✗ Statement (B) — v = u + ½at — the extra factor ½ is wrong; the correct form is v = u + at.
✗ (A) and (B) and (B) and (C) — both include a wrong equation, so neither combination can be correct.
📌 SSC & Railway Exam Facts
• The three equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• For free fall, replace a by g = 9.8 m/s² and take u = 0 when the body is simply dropped.
• Distance in the nth second: s(n) = u + (a/2)(2n − 1).
• These equations are valid only for uniform acceleration along a straight line.
• For a body dropped from rest, h = ½gt² and v = √(2gh).

Correct Answer: Option 4 — 40 m/s
📖 Given
• Initial velocity, u = 20 m/s
• Uniform acceleration, a = 4 m/s²
• Time, t = 5 s
✍️ Formula
v = u + at
🔢 Solution
• v = 20 + (4 × 5)
• v = 20 + 20
• v = 40 m/s
The train gains 20 m/s of extra speed in those 5 seconds, so its speed doubles from 20 m/s to 40 m/s.
❌ Why the other options are incorrect
✗ 60 m/s — this comes from multiplying, i.e. 20 + 4 × 10, or from taking t = 10 s instead of 5 s.
✗ 80 m/s — this is 20 × 4, i.e. the initial velocity multiplied by the acceleration; the formula is an addition, not a product.
✗ 16 m/s — this is 20 − 4; the train is accelerating, so the velocity must increase, not fall.
📌 SSC & Railway Exam Facts
• v = u + at is the first equation of motion and gives velocity at any time t.
• Use s = ut + ½at² for distance and v² = u² + 2as when time is not given.
• In a velocity–time graph, the slope gives acceleration and the area under the graph gives displacement.
• Uniform acceleration means the velocity changes by equal amounts in equal intervals of time.
• SI unit of velocity is m/s and of acceleration m/s²; both are vector quantities.

Correct Answer: Option 3 — non-uniform acceleration
Acceleration is the rate of change of velocity with time, a = (v − u)/t. Its SI unit is m/s² and it is a vector quantity.
Acceleration is said to be uniform when the velocity changes by equal amounts in equal intervals of time, and non-uniform (variable) when the velocity changes by unequal amounts in equal intervals of time. In this question the car's speed rises by a different amount in every equal time interval, so the rate of change of velocity is itself changing — the car moves with non-uniform acceleration.
A car being driven through crowded city traffic, where the driver keeps pressing and releasing the accelerator, is the standard everyday example.
❌ Why the other options are incorrect
✗ uniform speed — this means the car covers equal distances in equal intervals of time; its speed would not change at all.
✗ uniform velocity — velocity constant in both magnitude and direction; acceleration would then be zero.
✗ uniform acceleration — this needs the speed to increase by equal amounts in equal intervals, e.g. a freely falling body.
📌 SSC & Railway Exam Facts
• Speed is a scalar, velocity is a vector; both have SI unit m/s.
• a = (v − u)/t; SI unit m/s².
• Free fall is the classic case of uniform acceleration, g = 9.8 m/s².
• Negative acceleration is called retardation or deceleration.
• For uniform acceleration the velocity–time graph is a straight line; for non-uniform acceleration it is a curve.

Correct Answer: Option 3 — v = 2π r/t
📖 Given
• Radius of the circular path = r
• Time taken for one complete round = t seconds
• Distance covered in one round = circumference of the circle
✍️ Formula
Speed = Total distance / Total time, and circumference of a circle = 2πr
🔢 Solution
• Distance covered in one round, s = 2πr
• Time taken, t = t seconds
• v = s/t
• v = 2πr/t
So an athlete running once round a circular track covers a length equal to the circumference 2πr, and dividing it by the time gives the uniform circular speed.
❌ Why the other options are incorrect
✗ v = π r² /t — πr² is the area of the circle, not its perimeter; area divided by time is not a speed.
✗ v = 2 m/s — a fixed numerical value cannot be the answer when r and t are unknown symbols.
✗ v = 0 — the displacement after one full round is zero, so the average velocity is zero, but the speed (distance/time) is not zero.
📌 SSC & Railway Exam Facts
• Circumference of a circle = 2πr; area = πr².
• In uniform circular motion the speed is constant but the velocity keeps changing because the direction changes continuously — hence it is an accelerated motion.
• The acceleration in circular motion is centripetal acceleration, a = v²/r, directed towards the centre.
• Distance is a scalar, displacement is a vector; in one full circle distance = 2πr but displacement = 0.
• Time for one complete revolution is called the time period (T); v = 2πr/T.

Correct Answer: Option 3 — Change in velocity per unit time - Acceleration
Acceleration is defined exactly as the change in velocity per unit time:
a = (v − u)/t, SI unit m/s², a vector quantity. So this pair is correctly matched.
The other three pairs are deliberately swapped. Uniform motion means the velocity does not change with time (equal distances in equal intervals), while non-uniform motion means the velocity keeps changing with time. A freely falling body falls under gravity alone, and since g = 9.8 m/s² is constant near the Earth's surface, its motion is a textbook case of uniform acceleration.
❌ Why the other options are incorrect
✗ Velocity does not change with time - Non-uniform motion — wrong pairing; unchanging velocity is uniform motion.
✗ Velocity varies with time - Uniform motion — wrong pairing; varying velocity is non-uniform (accelerated) motion.
✗ Motion of a freely falling body - Non-uniform acceleration — a freely falling body has uniform acceleration g, so the pairing is wrong.
📌 SSC & Railway Exam Facts
• v = u + at, s = ut + ½at², v² = u² + 2as are the three equations of motion for uniform acceleration.
• In uniform motion, average speed = instantaneous speed and the distance–time graph is a straight line.
• g = 9.8 m/s² (often taken as 10 m/s²) and is independent of the mass of the falling body.
• For a body thrown up, g is negative (retardation); while coming down it is positive.
• Slope of a velocity–time graph = acceleration; area under it = distance travelled.

Correct Answer: Option 3 — 0 m
📖 Given
• First motion: 15 m to the right of point A
• Second motion: 15 m in the opposite direction
• Final position = the same point A (the starting point)
✍️ Formula
Displacement = shortest distance between the initial and the final position (a vector)
Distance = total path length covered (a scalar)
🔢 Solution
• Take rightward as positive: first displacement = +15 m
• Leftward (opposite) displacement = −15 m
• Net displacement = (+15) + (−15)
• Displacement = 0 m
• Total distance covered = 15 + 15 = 30 m
Because the object finishes exactly where it started, its displacement is zero even though it has actually travelled 30 m.
❌ Why the other options are incorrect
✗ 30 m — this is the total distance (path length), obtained by adding both legs without their directions.
✗ 15 m — this is only one leg of the journey, not the net change of position.
✗ 7.5 m — this comes from wrongly averaging the two legs; displacement is never an average of distances.
📌 SSC & Railway Exam Facts
• Displacement ≤ Distance always; they are equal only for motion in a straight line without reversing.
• Displacement can be zero, positive or negative; distance is never negative.
• SI unit of both is the metre (m).
• In one complete circular round, distance = 2πr and displacement = 0.
• Average speed = total distance/total time, while average velocity = total displacement/total time — here the average velocity is 0.

Correct Answer: Option 3 — 10 m/s
📖 Given
• Initial velocity, u = 0 m/s (the car starts from rest)
• Uniform acceleration, a = 0.5 m/s²
• Time, t = 20 s
• To find: final velocity, v
✍️ Formula
v = u + at (first equation of motion)
🔢 Solution
• v = 0 + (0.5 × 20)
• v = 0 + 10
• v = 10 m/s
In plain language: the car gains 0.5 m/s of speed every second, so in 20 s it gains 0.5 × 20 = 10 m/s, and since it began from rest that gain is its final velocity.
❌ Why the other options are incorrect
✗ 5 m/s — this is what you get if you take t = 10 s instead of 20 s (0.5 × 10).
✗ 0 m/s — this is the initial velocity u, not the velocity after 20 s.
✗ 15 m/s — this comes from wrongly using a = 0.75 m/s² or t = 30 s; it does not follow from the given data.
📌 SSC & Railway Exam Facts
• The three equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• Distance covered here would be s = ut + ½at² = 0 + ½ × 0.5 × 400 = 100 m.
• SI unit of acceleration is m/s²; of velocity, m/s.
• Slope of a velocity–time graph = acceleration; area under it = distance.
• 1 m/s = 3.6 km/h, so 10 m/s = 36 km/h.

Correct Answer: Option 3 — 15 m
📖 Given
• Distance covered, s₁ = 25 m
• Time taken, t₁ = 5 s
• Motion is uniform, so the speed stays constant
• To find: distance s₂ covered in t₂ = 3 s
✍️ Formula
Speed = Distance / Time, i.e. v = s/t, and then s = v × t
🔢 Solution
• v = 25/5
• v = 5 m/s
• s₂ = v × t₂ = 5 × 3
• s₂ = 15 m
In plain language: Rambabu walks 5 metres every second, so in 3 seconds he will cover 5 × 3 = 15 m.
❌ Why the other options are incorrect
✗ 5 m — this is his speed in m/s (5 m/s), mistakenly written as the distance.
✗ 20 m — this comes from subtracting wrongly (25 − 5) instead of using speed × time.
✗ 10 m — this is what you get with t = 2 s (5 × 2), not 3 s.
📌 SSC & Railway Exam Facts
• In uniform motion a body covers equal distances in equal intervals of time, so acceleration = 0.
• Speed = distance/time; SI unit m/s, and 1 m/s = 3.6 km/h — so 5 m/s = 18 km/h.
• For uniform motion the distance–time graph is a straight line whose slope gives the speed.
• Average speed = total distance/total time; for uniform motion it equals the instantaneous speed.
• Distance is a scalar and displacement a vector; in straight-line motion without reversal they are equal.

Correct Answer: Option 4 — A satellite in a circular orbit around the earth.
Uniform circular motion is the motion of a body along a circular path with a constant speed. Its speed does not change, but because the direction of motion changes continuously, the velocity keeps changing — so uniform circular motion is an accelerated motion.
A satellite revolving in a circular orbit around the Earth covers equal arcs in equal times at a constant speed while the gravitational pull of the Earth acts as the centripetal force that keeps bending it into a circle. This is the textbook example of uniform circular motion.
The acceleration in such motion is the centripetal acceleration a = v²/r, always directed towards the centre, and the speed of one revolution is v = 2πr/T.
❌ Why the other options are incorrect
✗ A person walking on a sidewalk — this is motion along a straight line (rectilinear motion), not along a circle.
✗ A car driving straight on a highway — again a straight-line motion; there is no continuous change of direction.
✗ A bird flying in the sky — the bird's path is random and irregular, and its speed also varies, so it is neither circular nor uniform.
📌 SSC & Railway Exam Facts
• In uniform circular motion the speed is constant but the velocity is variable, so it is an accelerated motion.
• Centripetal force F = mv²/r acts towards the centre; the outward "centrifugal force" is a pseudo force.
• v = 2πr/T, where T is the time period of one revolution.
• Other examples: the Moon around the Earth, the tip of a clock's second hand, a stone whirled on a string.
• If the string breaks, the stone flies off along the tangent to the circle — an example of inertia of direction.

Correct Answer: Option 3 — time = distance/speed
When a body is not accelerating, its speed remains constant — this is uniform motion, in which the body covers equal distances in equal intervals of time.
For such motion the basic relation is
Speed = Distance / Time, i.e. v = s/t.
Rearranging this single relation gives the two other correct forms:
Distance = Speed × Time and Time = Distance / Speed.
Only Option 3, time = distance/speed, is one of these correct rearrangements, so it is the answer. A quick unit check confirms it: metre ÷ (metre/second) = second, which is indeed a time.
❌ Why the other options are incorrect
✗ distance = time/speed — units give second ÷ (m/s) = s²/m, which is not a length; the correct form is distance = speed × time.
✗ distance = speed/time — units give (m/s) ÷ s = m/s², which is an acceleration, not a distance.
✗ speed = time/distance — this is the reciprocal of speed; its unit would be s/m, not m/s.
📌 SSC & Railway Exam Facts
• v = s/t, s = v × t, t = s/v — the same relation written three ways.
• SI unit of speed is m/s; 1 m/s = 3.6 km/h and 1 km/h = 5/18 m/s.
• In uniform motion acceleration = 0 and the distance–time graph is a straight sloping line, its slope giving the speed.
• In non-uniform motion the distance–time graph is a curve.
• Average speed = total distance / total time; for uniform motion it equals the instantaneous speed.

Correct Answer: Option 1 — (u + v)/2
Average velocity is defined as the total displacement divided by the total time taken. When a body moves with uniform acceleration, its velocity changes at a steady rate, so the average velocity over a time interval is simply the arithmetic mean of the initial and the final velocity:
v(av) = (u + v)/2
This works because for uniform acceleration the velocity–time graph is a straight line; the area under it (the displacement) equals the area of a trapezium, which is the mean height × base, i.e. ((u + v)/2) × t.
So, for example, a car that speeds up uniformly from 20 m/s to 40 m/s has an average velocity of (20 + 40)/2 = 30 m/s over that interval.
❌ Why the other options are incorrect
✗ u + v — this is the sum of the two velocities; it is always larger than the final velocity itself, which is impossible for an average.
✗ (u − v)/2 — this uses the difference of the velocities; (v − u) is related to change in velocity, which gives acceleration when divided by time, not average velocity.
✗ 0 — the average velocity is zero only when the net displacement is zero, for example after one complete circular round; it is not a general result.
📌 SSC & Railway Exam Facts
• Average velocity = total displacement/total time; average speed = total distance/total time.
• v(av) = (u + v)/2 holds only for uniformly accelerated motion.
• The three equations of motion: v = u + at, s = ut + ½at², v² = u² + 2as.
• Displacement can also be written as s = ((u + v)/2) × t.
• Velocity is a vector, speed a scalar; both have SI unit m/s.