Multiple choice

A car moving with a speed of $50$ km/hr can be stopped by applying brakes after $6$ m. If the same car is moving at a speed of $100$ km/hr, the minimum stopping distance is ? Assume constant retardation.

  1. 12 m

  2. 18 m

  3. 24 m

  4. 6 m
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C Correct answer
Explanation

Stopping distance is proportional to the square of the velocity (d = v^2 / 2a). Since the speed doubles from 50 km/hr to 100 km/hr, the stopping distance increases by a factor of 2^2 = 4. Thus, 6 m * 4 = 24 m.

AI explanation

By the third equation of motion, v^2 equals u^2 minus 2as. With a final velocity of 0, the stopping distance s is directly proportional to the square of the initial velocity u. When the speed increases from 50 km/hr to 100 km/hr, the initial velocity doubles, meaning the stopping distance increases by a factor of 2^2, which is 4. Multiplying the original stopping distance of 6 m by 4 gives a minimum stopping distance of 24 m. The minimum stopping distance is 24 m.

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