A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km $h^{- 1}$ to 18 km $h^{- 1}$ in 5 s. What is the distance travelled by the train during this interval of time?

52 m

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50 m

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25 m

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80 m

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Solution

The correct option is A 50 m
Initial velocity , $u = 54 k m / h = 54 (1000 / 3600) m / s = 15 m / s$ and final velocity, $v = 18 k m / h = 18 (1000 / 3600) = 5 m / s$
Using formula, $v = u - a t$ , $5 = 15 - a (5)$ or $a = 2 m / s^{2}$ , this is deacceleration of train.
Using, $S = u t - \frac{1}{2} a t^{2},$ the distance traveled by train, $d = 15 (5) - 0.5 (2) (5)^{2} = 50 m$

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A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km h−1 to 18 km h−1 in 5 s. What is the distance travelled by the train during this interval of time?

The correct option is A 50 mInitial velocity , u=54km/h=54(1000/3600)m/s=15m/s and final velocity, v=18km/h=18(1000/3600)=5m/sUsing formula, v=u−at, 5=15−a(5) or a=2m/s2, this is deacceleration of train.Using, S=ut−12at2, the distance traveled by train, d=15(5)−0.5(2)(5)2=50m

A driver applies brakes when he sees a child on the railway track, the speed of the train reduces from 54 km $h^{- 1}$ to 18 km $h^{- 1}$ in 5 s. What is the distance travelled by the train during this interval of time?