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)

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