Discussion Forum

The declaration of a car travelling on a straight highway is a function of its instantaneous velocity v given by $\omega=a\sqrt{v}$, where a is a constant. If the initial velocity of the car is 60 km/h, the distance of the car will travel and the time it takes before it stops  are 

Answer:

Explanation:

Deceleration $\omega=-a\sqrt{v}$

But   $\omega=\frac{dv}{dt}\Rightarrow\frac{dv}{dt}=-a\sqrt{v}$

$\frac{-dv}{\sqrt{v}}=a.dt\Rightarrow\int_{v_{i}}^{0} \frac{dv}{\sqrt{v}}=\int adt$

 $|2\sqrt{v}|_{v_{i}}^{0}=at\Rightarrow t=\frac{a}{2}\sqrt{v_{i}s}$

Again, 

$\frac{-dv}{dt}=a\sqrt{v}\Rightarrow\frac{dv}{dx}.\frac{dx}{dt}=-a\sqrt{v}$

$\frac{dv}{dx}.v=-a\sqrt{v}\Rightarrow dv \sqrt{v}=-a.dx$

 $\int_{v_{0}}^{0} \sqrt{v}dv=-a\int_{0}^{s} ds$

 After solving , we get

   $s=\frac{2}{3a}.v_{0}^{3/2}$