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, $u_1=u,v_1=0,t_1=t$
$\therefore v_1=u_1+a_{1}t0=u-a_{1}t$
$\therefore u=a_{1}t$
Now, $v_1^2=u_1^2+2a_1{s}_{1}0$
$=u^2-2a_1{s}_1{u}^2=2a_1{s}_1$
$\Rightarrow u^2=2\times \frac{u}{t}\times s_1$
$\therefore s_1=\frac{ut}{2}$ ...(ii)
For second car, $u_2=4u,v_2=0$
$\therefore v_2=u_2+at0=4u-a_{2}t$
$a_2=\frac{4u}{t}$ ...(iii)
Now, $v_2^2=u_2^2+2a_2{s}_{2}0$
$=u_2^2-2a_2{s}_2{u}_2^2=2a_2{s}_2$
$(4u{)}^2=2\times \frac{4u}{t}\times s_2$
$\therefore s_2=\frac{4ut}{2}$
$\therefore \frac{s_1}{s_2}=\frac{ut/2}{4ut/2}=\frac{1}{4}$