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Consider a car moving along a straight horizontal road with a speed of 72 km/h. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (taking g = 10 m/s²)
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- 30 m
- 40 m
- 72 m
- 20 m
Correct Option: B
Here u = 72 km/h = 20 m/s; v = 0;
a = – µg = – 0.5 × 10 = – 5 m/s²
As v² = u² + 2as,
| ∴ s = | = | = 40 m | ||
| 2a | 2 × (-5) |
