Question

At $t=0$, a body starts from origin with some initial velocity, The displacement $x\left(m\right)$ of the body varies with time $t\left(x\right)$ as $x=-\left(2/3\right)t^2+16t+2$. Find the initial velocity of the body and also find how long does the body take to come to rest? What is the acceleration of the body when it comes to rest?

Answer 1

$v=\frac{dx}{dt}=\frac{d}{dt}(-\frac{2}{3}{t}^3+16t+2)=-\frac{4}{3}t+16$
At $t=0$,
$v=-\frac{4}{3}\times 0+16=16m{s}^{-1}$
Putting $v=0$, we get $-\frac{4}{3}t+16=0\Rightarrow t=12s$
Hence, the body takes $12s$ to come to rest (momentarily).
Now $a=\frac{dv}{dt}=\frac{d}{dt}(-\frac{4}{3}t+16)=-\frac{4}{3}m{s}^{-2}$
We see that acceleration is constant, so when the body comes to rest, its acceleration is $-4/3m{s}^{-2}$.