A body moving with constant retardation 'a' losses $\frac{2}{3}$ of its initial velocity $v_{0}$ . What is the distance covered by the body in the time during which this occurs?

{(\frac{2 v_{0}}{3 a})}^{2}

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\frac{4 v_{0}}{9 a}

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\frac{2 v_{0}}{3 a}

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\frac{2 v_{0}^{2}}{9 a}

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Solution

The correct option is A ${(\frac{2 v_{0}}{3 a})}^{2}$
A body moving with constant retardation $^{'} a^{'}$ losses $\frac{2}{3} v_{0}$ .
Initial velocity, $u = v_{0}$
Final velocity, $v = v_{0} - \frac{2}{3} v_{0} = \frac{v_{0}}{3}$
From 3rd equation of motion,
$2 a s = v^{2} - u^{2}$
$2 a s = (\frac{v_{0}}{3})^{2} - (v_{0})^{2}$
$2 a s = \frac{8 v_{0}^{2}}{9}$
$s = \frac{4 v_{0}^{2}}{9 a}$
$s = (\frac{2 v_{0}}{3 a})^{2}$
The correct option is A.

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A body moving with constant retardation 'a' losses 23 of its initial velocity v0. What is the distance covered by the body in the time during which this occurs?

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A body moving with constant retardation 'a' losses $\frac{2}{3}$ of its initial velocity $v_{0}$ . What is the distance covered by the body in the time during which this occurs?