Question

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3v^{2/3}$. If at an initial point, the velocity is $8m/s$, then the distance travelled by the object before it stops is

• A. 2 m
• B. 4 m
• C. 6 m
• D. 8 m

Answer 1

Answer: (B)

Given that, $a=$ deceleration $=-3v^{2/3}m/s^2$
At initial point $(t=0)$ velocity $=8m/s$
Distance travelled by the object before it stop is We know that,
$a=\frac{dv}{dt}$
$\Rightarrow a=\frac{dv}{ds}\cdot \frac{ds}{dt}$
$a=vdv/ds$
$ads=vdv$
$-3v^{2/3}ds=vds$
$ds=-\frac{1}{3}{v}^{1/3}dv$
Integrate both sides, we get
$ds=\frac{-1}{3}\int v^{1/3}dv$
$\Rightarrow s=\frac{-1}{4}{v}^{4/3}+C\dots (i)$
$\Rightarrow$ At $t=0,u=8m/s,s=0$
So, $C=\frac{1}{4}(8{)}^{4/3}=\frac{16}{4}=4$
Therefore, from Eq. (i)
$s=\frac{-1}{4}{v}^{4/3}+4$
Body stops, when $v=0$
So, $s=4m$