Multiple choice

Consider a car moving along a straight horizontal road with a speed of $72$ km/h. If the coefficient of static friction between the tyre and the road is $0.5$, the shortest distance in which the car can be stopped is?(Take $g=10 ms^{-2}$)

  1. $30$m
  2. $40$m
  3. $72$m
  4. $20$m
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B Correct answer
Explanation

The stopping distance d is given by v^2 / (2 * mu * g). Convert 72 km/h to 20 m/s. d = (20^2) / (2 * 0.5 * 10) = 400 / 10 = 40 meters.

AI explanation

Using the equation of motion v squared equals u squared plus 2as, the final velocity v is zero and the initial velocity u is 72 km/h, which equals 20 m/s. The maximum deceleration is the coefficient of static friction multiplied by g, giving 0.5 times 10, which equals 5 m/s squared. Setting up the equation gives zero equals 20 squared minus 2 times 5 times s, meaning 400 equals 10s. Solving for s yields a stopping distance of 40 meters.

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