Question

A body of mass $2\, kg$ travels according to the law $x\left(t\right)$ $=pt+qt^{2}+rt^{3}$ where $p=3 \,ms^{-1}$, $q=4\,ms^{-2}$ and $r=5 \,ms^{-3}$. The force acting on the body at $t=2$ seconds is

• A. $136\,N$
• B. $134\, N$
• C. $158\, N$
• D. $68\, N$

Answer 1

Answer: (A)

Here, $x\left(t\right)$ $=pt+qt^{2}+rt^{3}$
where $p=3 \,m\,s^{-1}$, $q=4 \,m \,s^{-2}$, $r=5 \,m\, s^{-3}$ and $m=2 \,kg$
Velocity, $v=\frac{d\,x}{d \,t}=\frac{d}{d \,t}$ $\left(pt+qt^{2}+rt^{3}\right)=p+2qt+3rt^{2}$
Acceleration, $a=\frac{d\,v}{d \,t}$ $=2q+6rt$
At $t=2 s$, $a=2$ $\left(4 \,m\, s^{-2}\right)$ $+6\left(5 \,m \,s^{-3}\right)\left(2 s\right)$ $=8 \,m \,s^{-2}+60\, m\,s^{-2}=68 \,m\, s^{-2}$
The force acting on the body of mass $2 \,kg $ is
$F=ma$ $=\left(2 kg\right)\left(68 m s^{-2}\right)$ $=136 \,N$