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Different question variety formed

Question

A driver applies the brakes on seeing traffic signal 400 m ahead. At the time of applying the brakes, vehicle was moving with 15 ms^-1 and retarding with 0.3 ms^-2. The distance of vehicle after 1 minute from the traffic light is

Easy

A

25 m
B

375 m
C

360 m
D

40 m

Solution

Stopping distance $= \frac{v^{2}}{2 a}$
$= \frac{15 \times 15}{2 \times 3} = 375 m$
distance from the signal 400 - 375 = 25m

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