A car is moving with speed of 150 km / h and after applying the break it will move 27 m 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.

Solution

Physics sir · 3 steps

Same brakes, same deceleration: stopping distance goes as u², so a third of the speed gives a ninth.

1. Same deceleration. It is the same car with the same brakes, so the retardation a is the same in both cases. The final speed is zero each time.

   0 = u² - 2as ⇒ s = u²/2a ∝ u²

2. Take the ratio. s₂/s₁ = (u₂/u₁)² = (1/3)² = 1/9. There is no need to convert 150 km h⁻¹ to m s⁻¹, because only the ratio of speeds matters.
3. So. s₂ = 27/9 = 3 m. The answer is 3.

The common slip: 9. Treating stopping distance as proportional to speed itself gives 27 × 1/3 = 9 m. It goes as u², so divide by 9, not 3.

The idea to keep. For a fixed braking deceleration, stopping distance is proportional to the square of the initial speed. Work with the ratio and skip unit conversions.

Answer from the NTA final answer key. Solution by PrepWiser.