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 ${{v}_{0}}$​, the distance the car will travel and the time it takes before it stops are

$\begin{align}
  & \text{A}\text{. }\dfrac{2}{3}m\mathop{v}_{0}^{{}^{3}/{}_{2}},\dfrac{1}{2}\mathop{v}_{0}^{{}^{3}/{}_{2}}s \\
 & \text{B}\text{. }\dfrac{3}{2a}m\mathop{v}_{0}^{{}^{1}/{}_{2}},\dfrac{1}{2a}\mathop{v}_{0}^{{}^{1}/{}_{2}}s \\
 & \text{C}\text{. }\dfrac{3a}{2}m\mathop{v}_{0}^{{}^{1}/{}_{2}},\dfrac{a}{2}\mathop{v}_{0}^{{}^{3}/{}_{2}}s \\
 & \text{D}\text{. }\dfrac{2}{3a}m\mathop{v}_{0}^{{}^{3}/{}_{2}},\dfrac{2}{a}\mathop{v}_{0}^{{}^{1}/{}_{2}}s \\
\end{align}$

Answer 1

Hint: At first we are taking the deceleration of the car, and we are integrating it from the initial velocity to 0, and we are taking the distance from 0 to s, on solving the equation we will get the result required for the first point that is distance travelled after is decelerated. Now for the time taken during deceleration we will have to integrate velocity from initial to 0 and time from 0 to t second to get the required time.

Formula used:
$w=\dfrac{-vdv}{ds}$
$w=-\dfrac{dv}{dt}$

Complete step-by-step answer:
In the question it is given that the deceleration of the car travelling on a straight highway is a function of its instantaneous velocity which is given by:
$w=a\sqrt{v}$
Now we know that deceleration of a car can be written as,
$w=\dfrac{-vdv}{ds}$
Now we are integrating the velocity from ${{v}_{0}}$to 0 as the car is decelerating, and integrating the distance from 0 to s,
$\int\limits_{{{v}_{0}}}^{0}{{{v}^{{1}/{2}\;}}}dv=-a\int\limits_{0}^{s}{ds}$,
We get the result as,
$\dfrac{2}{3}{{v}_{0}}^{{3}/{2}\;}=as$,
Therefore,
$s=\dfrac{2}{3a}{{v}_{0}}^{{3}/{2}\;}$,this is the distance travelled by the car after deceleration starts and ends.
Now, using $w=-\dfrac{dv}{dt}$
Therefore,$-\dfrac{dv}{dt}=a\sqrt{v}$,
Now again integrating the velocity from initial to 0 and the time as 0 to t seconds,
We get $\int\limits_{{{v}_{0}}}^{0}{{{v}^{{-1}/{2}\;}}}dv=-a\int\limits_{0}^{t}{dt}$ ,
On solving this,
$2\sqrt{{{v}_{0}}}=at$ ,
So,
$t=\dfrac{2\sqrt{{{v}_{0}}}}{a}$

So, the correct answer is “Option D”.

Note: In the equation $\int\limits_{{{v}_{0}}}^{0}{{{v}^{{1}/{2}\;}}}dv=-a\int\limits_{0}^{s}{ds}$, we are integrating velocity from initial velocity to 0 as the car is decelerating its speed so at its end it will stop. And we are equating distance from 0 to ‘s’ as we are calculating the distance travelled from the beginning of deceleration to the end. In the second equation that is $\int\limits_{{{v}_{0}}}^{0}{{{v}^{{1}/{2}\;}}}dv=-a\int\limits_{0}^{s}{ds}$, we are calculating velocity in the same way as it is decelerating and the time of deceleration is calculated so we are taking the time as 0 to ‘t’.