Question

a body moving in a straight line with constant retardation "a" loses 3/4 th of its initial velocity U0 what is its time of motion

Answer 1

Answer:

The time of the motion is $\dfrac{3u_{0}}{4a}$

Explanation:

A body moving in a straight line with constant retardation "a" loses 3/4 th of its initial velocity u₀.

Initial velocity u= u₀

Final velocity $v = 1-\dfrac{3}{4}u_{0}$

Using equation of motion

$v= u+at$

The time of motion is

$t= \dfrac{v-u}{a}$

$t=\dfrac{\dfrac{1}{4}u_{0}-u_{0}}{-a}$

$t= \dfrac{3u_{0}}{4a}$

Hence, The time of the motion is $\dfrac{3u_{0}}{4a}$