Question

A particle moves along a straight line $OX$. At a time $t$ (in second), the distance $x$ (in metre) of the particle from $O$ is given by $x=40+12t-t^3$. How long would the particle travel before coming to rest :-

• A. 16 m
• B. 24 m
• C. 40 m
• D. 56 m

Answer 1

Answer: (A)

$x=40+12t-t^3$
$\therefore$ Velocity $v=\frac{dx}{dt}=12-3t^2$
When particle come to rest, $dx/dt=v=0$
$\therefore 12-3t^2=0$
$\Rightarrow 3t^2=12$
$\Rightarrow t=2sec$.
Distance travelled by the particle before coming to rest
$\underset{0}{\overset{s}{\int }}ds=\underset{0}{\overset{2}{\int }}vdt$
$s=\underset{0}{\overset{2}{\int }}\left(12-3t^2\right)dt=12t-{\frac{3t^3}{3}|}_0^2$
$s=12\times 2-8=24-8=16m$