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Q.

The displacement of a particle is given by $x = (t - 2)^{2}$ where x is in metres and t in seconds. The distance covered by the particle in first four seconds is

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a

4m

b

8m

c

12m

d

16m

answer is B.

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Detailed Solution

$x = (t - 2)^{2} = t^{2} + 4 - 4 t$
$\begin{aligned} v = \frac{d x}{d t} = 2 t - 4 \\ a = \frac{d v}{d t} = 2 {ms}^{- 2} \end{aligned}$
$At t = 0 \sec \to u = - 4 {ms}^{- 1}$
$S = ut + \frac{1}{2} {at}^{2} = - 4 (4) + \frac{1}{2} (2) (16)$
$\Rightarrow S = 8 m$

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