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Consider a car moving along a straight horizantal road with a speed of 72 km/h. If the coefficient of static friction between road and tyres is 0.5, the shortest distance in which the car can be stopped is
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- 30 m
- 40 m
- 72 m
- 20 m
Correct Option: B
Force due to friction = kinetic energy
| μ mg s = | mv2 | |
| 2 |
[ Here, v = 72 km/h
| = | = 20 m/s ] | |
| 60 × 60 |
| or , s = | = | = 40 m | ||
| 2μg | 2 × 0.5 × 10 |
