A car moving along straight road with a speed of 126 $k m h^{- 1}$ ( kmph) is brought to rest within a distance of 200 m. What is its retardation ? (assumed to be uniform) and also calculate the time taken for the car to stop.

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Solution

Initial velocity $u = 126 k m / h = 126 \times \frac{5}{18} = 35 m / s$
Final velocity $v = 0 m / s$
Distance covered $S = 200 m$
Using $v^{2} - u^{2} = 2 a S$
Or $0 - (35)^{2} = 2 \times a \times (200)$
$⟹ a = - 3 m / s^{2}$
Using $v = u + a t$
Or $0 = 35 + (- 3) t$
$⟹ t = 11.67 s$

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A car moving along straight road with a speed of 126 kmh−1 ( kmph) is brought to rest within a distance of 200 m. What is its retardation ? (assumed to be uniform) and also calculate the time taken for the car to stop.

Initial velocity u=126 km/h=126×518=35 m/sFinal velocity v=0 m/sDistance covered S=200 mUsing v2−u2=2aSOr 0−(35)2=2×a×(200)⟹ a=−3 m/s2Using v=u+atOr 0=35+(−3)t⟹ t=11.67 s

A car moving along straight road with a speed of 126 $k m h^{- 1}$ ( kmph) is brought to rest within a distance of 200 m. What is its retardation ? (assumed to be uniform) and also calculate the time taken for the car to stop.

Initial velocity $u = 126 k m / h = 126 \times \frac{5}{18} = 35 m / s$
Final velocity $v = 0 m / s$
Distance covered $S = 200 m$
Using $v^{2} - u^{2} = 2 a S$
Or $0 - (35)^{2} = 2 \times a \times (200)$
$⟹ a = - 3 m / s^{2}$
Using $v = u + a t$
Or $0 = 35 + (- 3) t$
$⟹ t = 11.67 s$