Multiple choice

A car is moving on a straight horizontal road with a speed of $72kmph$. If the coefficient of kinetic friction between the tyre of the car and the road is $0.5$, then find the minimum distance, within which the car can be stopped.

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

Speed u = 72 km/h = 20 m/s. Final speed v = 0. Using v^2 = u^2 + 2as, where a = -mu*g = -0.5 * 10 = -5 m/s^2. 0 = 20^2 + 2(-5)s, so 10s = 400, s = 40 m.

AI explanation

Using the kinematic equation v squared equals u squared plus 2as, the final velocity v is zero and the initial velocity u is 72 kmph, which equals 20 m/s. The maximum deceleration provided by kinetic friction is the coefficient of 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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