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 $60\, km/hr$, the distance the car will travel and the time it takes before it stops are

• A. $\frac{2}{3} m , \frac{1}{2}s $
• B. $\frac{3}{2a} m , \frac{1}{2a }s $
• C. $\frac{3a}{2} m , \frac{a}{2}s $
• D. non of these

Answer 1

Answer: (D)

Deceleration $\omega=-a \sqrt{V}$
But $\omega =\frac{d v}{d t} $
$\Rightarrow \frac{d V}{d t}=-a \sqrt{V} $
$\frac{-d v}{\sqrt{V}} =a \cdot d t$
$ \Rightarrow \int_\limits{v_{i}}^{0} \frac{d v}{\sqrt{V}}=\int a d t$
$|2 \sqrt{v}|_{v_{i}}^{0}=a t $
$ \Rightarrow t=\frac{2}{a} \sqrt{v_{i} S}$
Again, $\frac{-d v}{d t}=a \sqrt{v}$
$ \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=-a \sqrt{v} $
$\frac{d v}{d x} \cdot V=-a \sqrt{V} $
$\Rightarrow d v \sqrt{v}=-a \cdot d x$
$\int_\limits{v_{0}}^{0} \sqrt{v} d v=-a \int_\limits{0}^{s} d s$
After solving, we get
$S=\frac{2}{3 a} \cdot v_{0^{2}}$