Question

A particle of mass $2 / 3\, kg$ with velocity $v=-15 \,m / s$ at $t=-2 \,s$ is acted upon by a force $F=k-\beta t^{2}$. Here, $k=8\, N$ and $\beta=2 \,N / s ^{2}$. 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 \int\limits_{v=-15}^{v(t=2 s )} d v =\int\limits_{t=-2 s }^{t=2 s } \frac{k-\beta t^{2}}{m} d t =\frac{3}{2} \int\limits_{t=-2}^{t=2}\left(8-2 t^{2}\right) d t$
$v = \left(t=2s\right)-\left(-15\right)=\frac{3}{2}\left[8t-\frac{2t^{3}}{3}\right]_{-2}^{2}$
$\Rightarrow v_{\text{at}\, t=2\,s}=-15+\frac{3}{2}\times$
$\left[8\left(2-\left(-2\right)\right)-\frac{2}{3}\left(8\left(-2\right)^{3}\right)\right]$
$\Rightarrow v_{\text{at}\, t=2\,s}=-15+\frac{3}{2}\left(32-\frac{32}{3}\right)$
$=-15+\frac{3}{2}\left(\frac{64}{3}\right)$
$=-15+32=17 \,ms^{-1}$