Multiple choice

A car moving with a speed of $40 km/hr$ can be stopped by applying breaks for $2$ km. If the same car is moving with a speed of $80 km/hr$, what is the minimum stopping distance? Assume constant retardation.

  1. $8 Km$
  2. $2 Km$
  3. $4 Km$
  4. $6 Km$
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A Correct answer
Explanation

Stopping distance d is proportional to the square of the initial speed v^2 (since v^2 = 2ad). If speed doubles from 40 to 80, the stopping distance increases by a factor of 2^2 = 4. Thus, 2 km * 4 = 8 km.

AI explanation

When a vehicle stops under constant retardation, the stopping distance is proportional to the square of its initial velocity. Using the ratio D2 / D1 = (V2 / V1)^2, we substitute the given values to find D2 / 2 = (80 / 40)^2. This simplifies to D2 / 2 = 4, meaning the minimum stopping distance is 8 km.

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