Question

speed of two identical cars are u and 4u at a specific instant,the ratio of the respective distance in which the two cars are stopped from that instant by same breaking force-

Answer 1

Answer : $\frac{1}{16}$

Solution:

By Newtons third law of motion
$v^{2}-u^{2}=2as$

here the acceleration a is negative since it is retardation, displacement s is the stopping distance d, u is the initial velocity and v is the final velocity.
Since the objects stops after the motion, v=0
That is
  $-u^{2} = -2ad$
   ⇒$d= \frac{u}{2a}$

Lets take $d_{1}$ as the distance traveled by the car with velocity $u$ and $d_{2}$ as the distance traveled by the car with velocity  $4u$

The retardation for both the cars is -a
   ⇒$\frac{d_{1}}{d_{2}}= \frac{u^{2}}{2a} \frac{2a}{(4u)^{2}}$
   ⇒$\frac{d_{1}}{d_{2}}=\frac{1}{16}$

Answer 2

please mark as brain list.