Exams › NEET › Physics

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):

1. 30 m
2. 40 m
3. 72 m
4. 20 m

Correct answer: 40 m

Solution

The car's initial speed is 72 km/h = 20 m/s. The maximum deceleration is given by friction, a = μg = 0.5 × 10 = 5 m/s². Using the equation v² = u² - 2as, where v = 0, u = 20 m/s, and a = 5 m/s², we get s = u² / (2a) = 20² / (2 × 5) = 40 m.

Related NEET Physics questions

• In the equation of motion, \( \boldsymbol{S}=\boldsymbol{u} \boldsymbol{t}+ \) \( 1 / 2 a t^{2}, \) S stands for
• A body is accelerated by applying a force of 30 N. The change in the momentum of the body after 2 sec is?
• A passenger in moving train tosses a coin which falls behind him. It means that the motion of the train is
• From the displacement-time graph shown here, find the velocity of the body as it moves from \( A \) to \( B \)
• The distance of a particle as a function of time is shown below. The graph indicates that
• Whenever an object moves with a constant speed, its distance - time graph is a

⚔️ Practice NEET Physics free + battle 1v1 →