A body starting from rest, accelerates at a constant rate a $m/s^{2}$ for some time after which it decelerates at a constant rate $b\ m/s^{2}$ to come to rest finally. If the total time elapsed is $t\ sec$, the maximum velocity attained by the body is given by

Question

Dr. Abubacker · Accepted Answer

(a) Total time of motion $= t$
Duration of acceleration $= t'$
Duration of deceleration $= t - t'$
Given $u = 0$, $a =$ constant acceleration and
$b =$ constant deceleration
$v = 0 + at'$
Also $0 = v - b\left( t - t^{'} ight)$
$	herefore v = at'$
From (ii), $- v = - bt + bt'$
$\Rightarrow - at^{'} = - bt + bt'$
$\Rightarrow (a + b)t^{'} = bt \Rightarrow t^{'} = \frac{b}{(a + b)}t$
But $v = at'$
$	herefore$ Maximum velocity attained $= at'$
$\Rightarrow v = \frac{ab}{(a + b)}t\ m/s$

Q. A body starting from rest, accelerates at a constant rate a $m/s^{2}$ for some time after which it decelerates at a constant rate $b\ m/s^{2}$ to come to rest finally. If the total time elapsed is $t\ sec$, the maximum velocity attained by the body is given by

Explanation

Reviewed by Dr. Abubacker

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