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A body of mass 0.4 kg is moving initially with a constant speed of 10 ms^-1 to north is subjected to a constant force of 8 N directed towards south. At t = 0 if it is at x =0, then its position at t = 5 s is

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−200 m

zero

−250 m

answer is A.

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Detailed Solution

$S = u t + \frac{1}{2} a t^{2}$

$\begin{array}{l} x = 10 \times 5 + \frac{1}{2} (\frac{- 8}{0.4}) \times 5^{2} \\ = 50 - 250 m = - 200 m \end{array}$

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