Q158 / 160 · TS EAMCET 2017 · Physics

The deceleration of a car traveling 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 $60 \mathrm{~km} / \mathrm{h}$, the distance of the car will travel and the time it takes before it stops are

  • A
    $\frac{2}{3} m_1 \frac{1}{2} s$
  • B
    $\frac{3}{2 a} m, \frac{1}{2 a} s$
  • C
    $\frac{3 a}{2} m, \frac{a}{2} s$
  • Correct answer
    D
    $\frac{2}{3 a} m, \frac{2}{a} s$

Solution

Deceleration $\omega=-a \sqrt{v}$. But $\quad \omega=\frac{d v}{d t} \quad \Rightarrow \frac{d v}{d t}=-a \sqrt{v}$ $$ \begin{gathered} \frac{-d v}{\sqrt{V}}=a \cdot d t \Rightarrow \int_{v_1}^0 \frac{d v}{\sqrt{V}}=\int a d t \\ |2 \sqrt{v}|_{v_1}^0=a t \quad \Rightarrow t=\frac{2}{a} \sqrt{v_3 S} \end{gathered} $$ Again, $$ \begin{aligned} \frac{-d v}{d t} & =a \sqrt{v} \Rightarrow \frac{d v}{d x} \cdot \frac{d x}{d t}=-a \sqrt{v} \\ \frac{d v}{d x} \cdot v & =-a \sqrt{v} \Rightarrow d v \sqrt{v}=-a \cdot d x \\ \int_{v_0}^0 \sqrt{v} d v & =-a \int_0^s d s \end{aligned} $$ After solving, we get $$ s=\frac{2}{3 a} \cdot v_0^{\frac{3}{2}} $$

Hence, the correct answer is option: 'D'.

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