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 road and tyres is 0.5, the shortest distance in which the car can be stopped is

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

Correct answer: 40 m

Solution

The shortest stopping distance is determined by the maximum deceleration due to static friction. The deceleration is given by \( a = \mu g \), where \( \mu = 0.5 \) and \( g = 9.8 \, \text{m/s}^2 \). Converting the speed to \( 20 \, \text{m/s} \), and using \( v^2 = u^2 - 2as \), we find \( s = 40 \, \text{m} \).

Related NEET Physics questions

• Conservation of momentum in a collision between particles can be understood from
• Assertion (A): According to Newton's third law sum of action and reaction is not equal to zero Reason (R) : The forces action and reaction acts on different bodies.
• Who proposed the concept of Inertia?
• Which of the following Newton's law(s) is/are not applicable in non-inertial reference frame.
• Assertion When a bus starts suddenly, a person standing in it falls backwards. Reason It is due to inertia of rest.
• A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is:

⚔️ Practice NEET Physics free + battle 1v1 →