Speeds of two identical cars are u and 4u at a sufficient instant. What is the ratio of the respective distance at which the two cars are stopped from that instant?

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Solution

By Newton's third law of motion
$v^{2} - u^{2} = 2 a s$
here 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 tops after the motion, $v = 0$
thus
$- u^{2} = - 2 a d$
$d = \frac{u}{2 a}$
Let 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 $4 u$ .
The retardation for both the car is $- a$
$\frac{d_{1}}{d_{2}} = \frac{u^{2}}{2 a} \frac{2 a}{(4 u)^{2}}$
$\frac{d_{1}}{d_{2}} = \frac{1}{16}$

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