Question

A particle $P$ moves along a straight line away from a fixed point $O$ obeying the relation $S=16+48t-t^3.$ The direction of $P$ after $t=4$ is

• A. $\overrightarrow{PO}$
• B. $\overrightarrow{OP}$
• C. Rest at the instant
• D. Perpendicular to $\overrightarrow{OP}$

Answer 1

The correct option is A $\overrightarrow{PO}$

Displacement is given by $S\left(t\right)=16+48t-t^3$
$V=\frac{dS}{dt}=48-3t^2$
Clearly, for $t>4,V<0$
Negative sign indicates that it is moving towards the origin.