Question

An object moves in a straight line with deceleration whose magnitude varies with velocity as $3 v^{2 / 3}$. If at an initial point, the velocity is $8 \,m / s$, 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 $=-3 v^{2 / 3} m / s ^{2}$
At initial point $(t=0)$ velocity $=8 \,m / s$
Distance travelled by the object before it stop is We know that,
$a =\frac{d v}{d t} $
$\Rightarrow \,a=\frac{d v}{d s} \cdot \frac{d s}{d t} $
$a =v d v / d s $
$a d s =v d v$
$-3 v^{2 / 3} d s =v d s $
$d s =-\frac{1}{3} v^{1 / 3} d v$
Integrate both sides, we get
$d s =\frac{-1}{3} \int v^{1 / 3} d v $
$\Rightarrow \, s =\frac{-1}{4} v^{4 / 3}+C \,\,\,\,\dots(i)$
$\Rightarrow $ At $t =0, u=8\, m / s , s=0$
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, $s=4 \,m$