A body moving in a straight line with constant retardation “a” loses 3/4th of its initial velocity .What is its time of motion?

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$\frac{3 u_{0}}{4}$

$\frac{3 u_{0}}{4 a}$

$\frac{4 u_{0}}{3}$

$\frac{4 u_{0}}{3 a}$

answer is B.

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Detailed Solution

$\begin{array}{l} u = u_{0} \\ v = u - \frac{3}{4} u_{0} \\ a = \frac{v - u}{t} \\ t = \frac{v - u}{a} \\ t = \frac{\frac{u_{0}}{4} - u_{0}}{a} \\ t = \frac{+ 3 u_{0}}{4 a} \end{array}$

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