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A car is moving on a horizontal road whose acceleration versus position graph is drawn below. At the initial moment of time when particle was at $x = 0$ , its speed was $\sqrt{10} m / s$ . Find the speed of the particle when it reaches at $x = 8 m$ .

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$\sqrt{110} m / s$

$\sqrt{84} m / s$

$10 m / s$

$11 m / s$

answer is B.

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Detailed Solution

Area under a versus x gives $\frac{v^{2} - u^{2}}{2}$

$\Rightarrow \frac{v^{2} - {(\sqrt{10})}^{2}}{2} = 45$

$∴ v = 10 m / s$

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