Physics · Kinematics · chapter 2 of 14

Class 11 Physics Chapter 2: Motion in a Straight Line case study question with answers

One case-based question of the kind CBSE-pattern school exams ask: a passage and a table, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

In a school laboratory, Meera places a motion sensor at one end of a level track and pushes a trolley so that it moves away from the sensor with a steady push. The sensor records the position of the trolley, measured from a mark on the track, once every second. Her readings are given in the table. She wants to know whether the acceleration of the trolley was uniform and, if so, its value.

t (s)	0	1	2	3	4
x (m)	0	1.5	4.0	7.5	12.0

Part 1 · 1 mark

The average velocity of the trolley between t = 0 and t = 4 s is

(a) 2 m s⁻¹(b) 3 m s⁻¹(c) 5 m s⁻¹(d) 12 m s⁻¹

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Answer: (b) 3 m s⁻¹

v̄ = Δx/Δt = (12.0 − 0)/4 = 3 m s⁻¹. 12 m is the displacement, not the velocity.

Part 2 · 1 mark

A position-time graph drawn from these readings is

(a) a straight line inclined to the time axis(b) a straight line parallel to the time axis(c) a curve whose slope increases with time(d) a curve whose slope decreases with time

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Answer: (c) a curve whose slope increases with time

The displacements in successive seconds, 1.5, 2.5, 3.5 and 4.5 m, keep increasing, so the slope (velocity) increases and the graph curves upward: positive acceleration.

Part 3 · 2 marks

Using the displacements in successive seconds, show that the acceleration was uniform, and find the acceleration and the velocity of the trolley at t = 0.

Answer and marking scheme

Displacements in the 1st to 4th seconds: 1.5, 2.5, 3.5, 4.5 m. They increase by the same 1.0 m every second, so the acceleration is uniform. Since sₙ = v₀ + (a/2)(2n − 1), successive values differ by a, so a = 1.0 m s⁻². For n = 1: 1.5 = v₀ + 0.5, so v₀ = 1.0 m s⁻¹.

equal increase of 1.0 m per second shows uniform acceleration, a = 1.0 m s⁻² (1)
v₀ = 1.0 m s⁻¹ from the first-second displacement (1)

Taking the acceleration as 1.0 m s⁻² and the velocity at t = 0 as 1.0 m s⁻¹ (as the readings show), find the velocity of the trolley at t = 4 s and predict its position at t = 6 s if the push continues unchanged.

Answer and marking scheme

v = v₀ + at = 1.0 + 1.0 × 4 = 5.0 m s⁻¹. x = x₀ + v₀t + ½at² = 0 + 1.0 × 6 + ½ × 1.0 × 36 = 24 m.

v = 5.0 m s⁻¹ at t = 4 s (1)
x = 24 m at t = 6 s (1)

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