Question:

A car is moving along a straight horizontal road with a speed \(v_0\). If the coefficient of friction between the tyres and the road is \(\mu\), the shortest distance in which the car can be stopped is

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Stopping distance is independent of mass of the car.
Updated On: Apr 23, 2026
  • \(\frac{v_0^2}{\mu g}\)
  • \(\left(\frac{v_0}{\mu g}\right)^2\)
  • \(\frac{v_0^2}{\mu g}\)
  • \(\frac{v_0^2}{2\mu g}\)
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The Correct Option is D

Solution and Explanation

Step 1: Understanding the Concept:
Work done by friction = Change in kinetic energy. Friction force \(f = \mu mg\).
Step 2: Detailed Explanation:
Initial KE = \(\frac{1}{2}mv_0^2\). Work done by friction = \(f \cdot s = \mu mg \cdot s\).
Equating: \(\frac{1}{2}mv_0^2 = \mu mg s \Rightarrow s = \frac{v_0^2}{2\mu g}\).
Step 3: Final Answer:
Thus, shortest stopping distance = \(\frac{v_0^2}{2\mu g}\).
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