Question

A car is moving with speed of 150km/h and after applying the brake it will move 27m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling ______m distance.

Answer 1

Correct Answer: 3

To solve this problem, we need to use the concept of stopping distance and the relationship between speed, stopping distance, and braking. The stopping distance \(d\) of a car is proportional to the square of its speed \(v\), assuming the braking force and other conditions remain constant, which means that \[d \propto v^2\].

We are given that when the car is moving at 150 km/h, it stops after traveling 27 m. We need to find the stopping distance when the speed is one third of 150 km/h, which is 50 km/h.

First, express the relationship as a ratio using the stopping distances and speeds:

\[\frac{d_1}{d_2} = \left(\frac{v_1}{v_2}\right)^2\]

Where:

• \(d_1 = 27\) m is the stopping distance at speed \(v_1 = 150\) km/h.
• \(d_2\) is the stopping distance at speed \(v_2 = 50\) km/h.

Substituting the known values, we have:

\[\frac{27}{d_2} = \left(\frac{150}{50}\right)^2 = 3^2 = 9\]

Thus,

\[d_2 = \frac{27}{9} = 3 \] m

Therefore, when the car is traveling at 50 km/h, it stops after traveling 3 m.

This computed distance of 3 m is validated against the provided range (min, max) 3, 3, confirming it fits within the expected values.