Question

Position (in $m$) of a particle moving on a straight line varies with time (in $sec$) as $x=(\frac{t^3}{3}-3t^2+8t+4$). Considering the motion of the particle from $t$ = 0 $sec$ to $t$ = 5 $sec$, let $S_1$ be the total distance travelled and $S_2$ be the distance travelled during retardation. If $\frac{S_1}{S_2}=\frac{(3\alpha +2)}{11}$, the value of $\alpha$ is. Write upto two digits after the decimal point.

Answer 1

Given,
$x=\frac{t^3}{3}-3t^2+8t+4$
On differentiating w.r.t time $t$ we get velocity equation,
$v=t^2-6t+8=(t-2)(t-4)$
Therefore velocity is zero at times $t$ equals to 2 $sec$ and 4 $sec$.
On again differentiating w.r.t time $t$ we get acceleration equation $a=2(t-3)$, i.e., at time $t=3sec$ the direction of velocity changes.


$S_1=|S_{0sec-2sec}|+|S_{2sec-4sec}|+|S_{4sec-5sec}|S_1=|(\frac{32}{3}-4)|+|(\frac{32}{3}-\frac{28}{3})|+|(\frac{32}{3}-\frac{28}{3})|=\frac{20}{3}+\frac{8}{3}=\frac{28}{3}m.S_2=|S_{0sec-2sec}|+|S_{2sec-3sec}|S_2=|(\frac{32}{3}-4)|+|(10-\frac{32}{3})|=\frac{20}{3}+\frac{2}{3}=\frac{22}{3}m\frac{S_1}{S_2}=\frac{28}{22}=\frac{14}{11}=\frac{3\alpha +2}{11}\Rightarrow \alpha =4$