A car moving at a speed of 30 kilo meters per hour’s is brought to a halt in 8 metres by applying brakes. If the same car is moving at 60 km. per hour, it can be brought to a halt with same braking power in

The correct answer is 32 metres.

This problem is based on the equations of motion under uniform retardation. When brakes are applied, the car decelerates uniformly. We use the third equation of motion:

v^2=u^2+2as

When the car is brought to a halt, the final velocity v=0. So the equation becomes:

0=u^2+2as
which gives us: s=\frac{u^2}{2\left|a\right|}

This tells us a key relationship: the stopping distance is directly proportional to the square of the initial speed, provided the braking force (and hence deceleration) remains the same.

s\propto u^2

Step 1: Set up the ratio.

Let s_1 = stopping distance at speed u_1=30 km/h = 8 metres
Let s_2 = stopping distance at speed u_2=60 km/h = ?

Using the proportionality:

\frac{s_2}{s_1}=\frac{u_2^2}{u_1^2}

Step 2: Substitute the known values.

\frac{s_2}{8}=\frac{{60}^2}{{30}^2}=\frac{3600}{900}=4

Step 3: Solve for s_2.

s_2=8\times 4=32 metres

Intuition: When the speed is doubled (from 30 to 60 km/h), the stopping distance increases by a factor of 2^2=4. So 8 × 4 = 32 metres. This is why speeding is so dangerous — even a small increase in speed leads to a dramatically longer braking distance.