A driver applies the brakes on seeing the red traffic signal 400 m ahead. At the time of applying brakes, the vehicle was moving with 15 m/s and retarding at 0.3 m/s². The distance of the vehicle from the traffic light one minute after application of brakes is (a) 375 m (b) 360 m (c) 40 m (d) 25 m

Answer: (d). Time to stop = 15/0.3 = 50 s, less than 60 s, so the vehicle is at rest after 1 minute. Stopping distance = u²/2a = (15²)/(2 × 0.3) = 375 m Distance from the light = 400 - 375 = 25 m So the answer is (d) 25 m.

A driver applies the brakes on seeing the red traffic signal 400 m… | MHT CET 2024 (2 May, Shift 2)

Q13MHT CET 2024 · 2 MAY · SHIFT 2Medium

Equations of Kinematics

A driver applies the brakes on seeing the red traffic signal 400 m ahead. At the time of applying brakes, the vehicle was moving with $15\text{ m/s}$ and retarding at $0.3\text{ m/s}^2$ . The distance of the vehicle from the traffic light one minute after application of brakes is

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Answer: (d)

Time to stop $= \frac{15}{0.3} = 50\text{ s}$ , less than 60 s, so the vehicle is at rest after 1 minute.
Stopping distance $= \frac{u^2}{2a} = \frac{15^2}{2 \times 0.3} = 375\text{ m}$
Distance from the light $= 400 - 375 = 25\text{ m}$
So the answer is (d) 25 m.

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Equations of Kinematics question from MHT CET 2024 (2 May, Shift 2)

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