Question

Speeds of two identical cars are $u$ are $4u$ at a specific instant. The ratio of the respective distances in which the two cars are stopped in the same time

• A. 1 : 1
• B. 1 : 4
• C. 1 : 8
• D. 1 : 16

Answer 1

Answer: (B)

For first car, $u_{1}=u, v_{1}=0, t_{1}=t$
$\therefore v_{1}=u_{1}+a_{1} t 0=u-a_{1} t$
$\therefore u=a_{1} t$
Now, $v_{1}^{2}=u_{1}^{2}+2 a_{1} s_{1} 0$
$=u^{2}-2 a_{1} s_{1} u^{2}=2 a_{1} s_{1}$
$\Rightarrow u^{2}=2 \times \frac{u}{t} \times s_{1}$
$\therefore s_{1}=\frac{u t}{2}$ ...(ii)
For second car, $u_{2}=4 u, v_{2}=0$
$\therefore v_{2}=u_{2}+a t 0=4 u-a_{2} t$
$a_{2}=\frac{4 u}{t}$ ...(iii)
Now, $v_{2}^{2}=u_{2}^{2}+2 a_{2} s_{2} 0$
$=u_{2}^{2}-2 a_{2} s_{2} u_{2}^{2} =2 a_{2} s_{2}$
$(4 u)^{2}=2 \times \frac{4 u}{t} \times s_{2}$
$\therefore s_{2}=\frac{4 u t}{2}$
$\therefore \frac{s_{1}}{s_{2}}=\frac{u t / 2}{4 u t / 2}=\frac{1}{4}$