Question

The deceleration of a car traveling on a straight highway is a function of its instantaneous velocity $v$ given by $w=a\sqrt{v}$, where $a$ is a constant. If the initial velocity of the car is $v_o$, the distance the car will travel and the time it takes before it stops are

• A. $\frac{2}{3}m{v}_o^{3/2},\frac{1}{2}{v}_o^{3/2}s$
• B. $\frac{3}{2a}m{v}_o^{1/2},\frac{1}{2a}{v}_o^{1/2}s$
• C. $\frac{3a}{2}m{v}_o^{1/2},\frac{a}{2}{v}_o^{3/2}s$
• D. $\frac{2}{3a}m{v}_o^{3/2},\frac{2}{a}{v}_o^{1/2}s$

Answer 1

The correct option is D $\frac{2}{3a}m{v}_o^{3/2},\frac{2}{a}{v}_o^{1/2}s$

Deceleration of the car is given as $w=a\sqrt{v}$
Using $w=\frac{-vdv}{ds}$
$\therefore$ ${\int }_{v_o}^o{v}^{1/2}dv=-a{\int }_o^{S}ds$
We get $\frac{2}{3}{v}_o^{3/2}=aS$
$⟹S=\frac{2}{3a}{v}_o^{3/2}$
Using $w=-\frac{dv}{dt}$
$\therefore$ $-\frac{dv}{dt}=a\sqrt{v}$
Integrating ${\int }_{v_o}^o{v}^{-1/2}dv=-a{\int }_o^{t}dt$
we get, $2\sqrt{v_o}=at$
$⟹t=\frac{2\sqrt{v_o}}{a}$