Question

A particle is travelling along a straight line $OX$. The distance $x$ (in metre) of the particle from $O$ at a time ${}^{\prime }{t}^{\prime }$ is given by $x=37+27t-t^3$, where 't' is time in second. The distance of the particle from $O$ ,when it comes to rest is:

• A. $81m$
• B. $91m$
• C. $101m$
• D. $111m$

Answer 1

Answer: (B)

$91m$

The position of particle is given as $x=37+27t-t^3$
Thus the velocity of particle $v=\frac{dx}{dt}=27-3t^2$

According to problem,
$v=0\Rightarrow 27-3t^2=0$
Here, we get $t=\sqrt{\frac{27}{3}}=\sqrt{9}=3s$
Thus the distance of particle when it comes to rest $x(t=3s)=37+27\times 3-(3{)}^3=91m$