Multiple choice

An automobile travelling with a speed of $60$ km/h can brake to stop in a distance of $20$ m. If the car is going twice as fast i.e., $120$ km/h the stopping distance will be (a ssume constant retardation).

  1. 20 m

  2. 40 m

  3. 60 m

  4. 80 m
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D Correct answer
Explanation

Stopping distance d is proportional to the square of the velocity (v^2 = 2ad). If velocity doubles, v^2 becomes 4 times larger, so the stopping distance becomes 4 * 20m = 80m.

AI explanation

Using the third equation of motion, v^2 equals u^2 minus 2as. Since the final velocity v is 0, the stopping distance s is proportional to the square of the initial velocity u. When the speed doubles from 60 km/h to 120 km/h, the initial velocity is multiplied by 2, so the stopping distance becomes 2^2, or 4, times larger. Multiplying the original stopping distance of 20 m by 4 gives a new stopping distance of 80 m. The stopping distance will be 80 m.

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