A car increases its speed from $20 k m / h r$ to $30 k m / h r$ in $10 s e c o n d s .$ Its acceleration is

30 m / s^{2}

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3 m / s^{2}

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18 m / s^{2}

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0.2778 m / s^{2}

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Solution

The correct option is D $0.2778 m / s^{2}$
Given $u = 20 k m / h r = \frac{50}{9} m / s$ , $v = 3 k m / h r = \frac{75}{9} m / s$
Using $v = u + a t$ $\Rightarrow \frac{75}{9} = \frac{25}{9} + a \times 10$
$\Rightarrow a = \frac{5}{18} = 0.2778$ $m / s^{2}$

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