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Question

If a car at rest accelerates uniformly to a speed of 144 km $h^{- 1}$ in 20 s, then it covers a distance of:
20 m
400 m
1440 m
2880 m

A

2880 m

B

20 m

C

400 m

D

1440 m

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Solution

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Given
$u = 0$
$v = 144 k m p h = 144 \times \frac{5}{18} = 40 m / s$
$t = 20 s e c$
We know,
$v = u + a t$
$40 = a \times 20$
$a = 2 m / s^{2}$
Now,
$s = u t + \frac{1}{2} a t^{2}$
$s = \frac{1}{2} a t^{2} = \frac{1}{2} \times 2 \times 20^{2} = 400 m$

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