Question

A particle of mass $2/3kg$ with velocity $v=-15m/s$ at $t=-2s$ is acted upon by a force $F=k-\beta t^2$. Here, $k=8N$ and $\beta =2N/s^2$. The motion is one-dimensional. Then, the speed at which the particle acceleration is zero again, is

• A. 1 m/s
• B. 16 m/s
• C. 17 m/s
• D. 32 m/s

Answer 1

Answer: (C)

Force on the object is $F=k-\beta t^2$

Acceleration of the particle,
$a=\frac{F}{m}=\frac{k-\beta t^2}{m}$
Acceleration is zero when $k=\beta t^2$
or $t^2=\frac{k}{\beta }$
or $t^2=\frac{8}{2}$
$\Rightarrow t=2s$
$Now,a=\frac{dv}{dt}=\frac{k-\beta t^2}{m}$
$\Rightarrow dv=\frac{k-\beta t^2}{m}.dt$
Integrating between given limits, we have
$\Rightarrow \underset{v=-15}{\overset{v(t=2s)}{\int }}dv=\underset{t=-2s}{\overset{t=2s}{\int }}\frac{k-\beta t^2}{m}dt=\frac{3}{2}\underset{t=-2}{\overset{t=2}{\int }}\left(8-2t^2\right)dt$
$v=\left(t=2s\right)-\left(-15\right)=\frac{3}{2}{\left[8t-\frac{2t^3}{3}\right]}_{-2}^2$
$\Rightarrow v_{\text{at}t=2s}=-15+\frac{3}{2}\times$
$\left[8\left(2-\left(-2\right)\right)-\frac{2}{3}\left(8{\left(-2\right)}^3\right)\right]$
$\Rightarrow v_{\text{at}t=2s}=-15+\frac{3}{2}\left(32-\frac{32}{3}\right)$
$=-15+\frac{3}{2}\left(\frac{64}{3}\right)$
$=-15+32=17m{s}^{-1}$