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The Vault
RowQ
The Vault
CBSE Class 11 Physics · 10 questions · 24 marks
Describing motion honestly means separating what a body did from where it ended up — that is the difference between distance and displacement, and between speed and velocity. Here you learn to read position-time and velocity-time graphs as fluently as equations, and to derive the kinematic equations using calculus rather than memorising them. Everything later in mechanics rests on getting one-dimensional motion completely under control.
A drone flies 40 m due east, then 30 m due west, all in 10 s. What are its average speed and the magnitude of its average velocity?
Answer
7 m/s and 1 m/s — total distance = 40 + 30 = 70 m, so average speed = 70/10 = 7 m/s. Net displacement = 40 − 30 = 10 m east, so the magnitude of average velocity = 10/10 = 1 m/s. The two differ because the motion reverses direction.
The position of a particle is given by x = 5t³ − 2t² + 7 (x in metres, t in seconds). What is its acceleration at t = 2 s?
Answer
56 m/s² — differentiating, v = dx/dt = 15t² − 4t and a = dv/dt = 30t − 4. At t = 2 s, a = 30(2) − 4 = 60 − 4 = 56 m/s².
A ball is thrown vertically upward with speed 19.6 m/s from ground level (g = 9.8 m/s²). What is the total time it spends in the air before returning to the thrower's hand?
Answer
4 s — the time to reach the highest point is t = u/g = 19.6/9.8 = 2 s, and the descent takes an equal 2 s because the motion is symmetric about the top. Total time of flight = 2u/g = 2(19.6)/9.8 = 4 s.
On a velocity-time graph, a straight line with negative slope lying entirely above the time axis represents a body that is:
Answer
Moving forwards while slowing down — the velocity stays positive (the line is above the axis) so the body keeps moving in the positive direction, while the negative slope means the acceleration is negative, so its speed decreases. This is retardation, not reversal.
Assertion (A): A body can have zero velocity at an instant and still have a non-zero acceleration at that instant. Reason (R): Acceleration is the rate of change of velocity and does not depend on the instantaneous value of velocity.
Answer
Both A and R are true and R is the correct explanation of A — a stone thrown straight up has v = 0 at the topmost point, yet a = g = 9.8 m/s² downward there. Since a = dv/dt depends on how fast velocity is changing rather than on its momentary value, a body momentarily at rest can still be accelerating.
A car moving at 25 m/s brakes uniformly and stops in 5 s. Find its retardation and the distance covered while stopping.
Answer
Using v = u + at with u = 25 m/s, v = 0, t = 5 s: 0 = 25 + a(5), so a = −5 m/s². The retardation is 5 m/s². Distance: v² = u² + 2as gives 0 = 625 + 2(−5)s, so s = 625/10 = 62.5 m.
A body starts from rest with uniform acceleration 3 m/s². Find the distance it covers in the 6th second, and explain why this is not the same as the distance covered in 6 seconds.
Answer
Distance in the nth second: s_n = u + (a/2)(2n − 1). With u = 0, a = 3 m/s², n = 6: s₆ = 0 + (3/2)(2 × 6 − 1) = (1.5)(11) = 16.5 m. Distance in 6 seconds: s = ut + ½at² = 0 + ½(3)(36) = 54 m. They differ because s₆ is the displacement during a single one-second interval, from t = 5 s to t = 6 s, whereas s is the total displacement accumulated over the whole 6 s. In fact s₆ = s(6) − s(5) = 54 − 37.5 = 16.5 m, which confirms the result.
Using calculus, derive the three equations of uniformly accelerated motion, v = u + at, s = ut + ½at² and v² = u² + 2as, stating clearly the assumption on which they rest.
Answer
Assumption: the acceleration a is constant in magnitude and direction throughout the motion, and the motion is along a straight line. Let the body have velocity u at t = 0 and velocity v at time t, covering displacement s. First equation: by definition a = dv/dt, so dv = a dt. Integrating from v = u at t = 0 to v = v at time t: ∫dv = a∫dt, giving v − u = at, hence v = u + at. Second equation: velocity v = ds/dt = u + at, so ds = (u + at)dt. Integrating from s = 0 at t = 0 to s at time t: s = ∫(u + at)dt = ut + ½at². Hence s = ut + ½at². Third equation: write a = dv/dt = (dv/ds)(ds/dt) = v(dv/ds). Then v dv = a ds. Integrating from v = u at s = 0 to v at displacement s: ∫v dv = a∫ds gives (v² − u²)/2 = as, hence v² = u² + 2as. Note that the third equation contains no time, which makes it the natural choice whenever the time of travel is neither given nor asked for.
A stone is dropped from the top of a tower. During the last second of its fall it covers 25 m. Taking g = 10 m/s², find (a) the total time of fall, (b) the height of the tower, and (c) the speed with which the stone strikes the ground.
Answer
(a) Let the total time of fall be n seconds, with u = 0 and a = g = 10 m/s². Distance covered in the nth second is s_n = u + (g/2)(2n − 1) = (10/2)(2n − 1) = 5(2n − 1). Given s_n = 25 m: 5(2n − 1) = 25, so 2n − 1 = 5, giving n = 3 s. The stone falls for 3 s. (b) Height h = ut + ½gt² = 0 + ½(10)(3)² = 5 × 9 = 45 m. The tower is 45 m tall. (c) Final speed v = u + gt = 0 + 10 × 3 = 30 m/s. As a check, v² = u² + 2gh = 0 + 2(10)(45) = 900, so v = 30 m/s downward, which agrees.
Read the following and answer the questions that follow: On a straight expressway, a delivery van travelling at a steady 20 m/s passes a stationary patrol car. Exactly 4 s later the patrol car starts from rest and accelerates uniformly at 4 m/s² along the same lane, keeping this acceleration until it draws level with the van. (a) Write expressions for the displacement of each vehicle measured from the patrol car's starting point, taking t as the time since the patrol car started. (b) Find how long after starting the patrol car catches the van. (c) How far from the starting point does the overtaking occur? (d) What is the patrol car's speed at that moment, and comment on it.
Answer
(a) In the 4 s head start the van already covers 20 × 4 = 80 m. Taking t as the time since the patrol car starts: Van: x_v = 80 + 20t (constant velocity, so no acceleration term). Patrol car: x_p = 0 + ½(4)t² = 2t². (b) The car catches the van when x_p = x_v: 2t² = 80 + 20t, i.e. 2t² − 20t − 80 = 0, or t² − 10t − 40 = 0. t = [10 ± √(100 + 160)]/2 = [10 ± √260]/2 = (10 ± 16.12)/2. Rejecting the negative root, t = 26.12/2 ≈ 13.1 s after the patrol car starts. (c) Distance x_p = 2t² = 2(13.1)² = 2 × 171.6 ≈ 343 m from the patrol car's starting point. Checking with the van: 80 + 20(13.1) = 80 + 262 = 342 m, which agrees to rounding. (d) Speed of the patrol car v = at = 4 × 13.1 ≈ 52.4 m/s. It is more than twice the van's speed, which is the point: because the pursuer starts from rest, it must build up to a far higher speed than the target in order to close both the head-start gap and the distance the target keeps adding.
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