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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) [1992]
- a)30 m
- b)40 m
- c)72 m
- d)20 m
Correct answer is option 'B'. Can you explain this answer?
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Consider a car moving along a straight horizontal road with a speed of...
Here u = 72 km/h = 20 m/s; v = 0;
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Consider a car moving along a straight horizontal road with a speed of...
Shortest Distance to Stop a Car Moving at 72 km/hour
Given:
To find: Shortest distance in which the car can be stopped
Solution:
f = µN
N = mg
f = µmg
F = ma
f = ma = mv²/r
f = ma = mv²/∞ = 0
F = ma = -µmg
a = -µg
t = v/a = -v/µg
S = vt + 1/2 at² = v²/2µg
Given:
- Speed of the car, v = 72 km/h = 20 m/s
- Coefficient of static friction between the tyres and the road, µ = 0.5
- Acceleration due to gravity, g = 10 m/s²
To find: Shortest distance in which the car can be stopped
Solution:
- The force that provides the retarding motion to the car is the frictional force between the tyres and the road. This force is given by:
- Where N is the normal force acting on the car.
- The normal force acting on the car is equal to the weight of the car which is given by:
- Where m is the mass of the car and g is the acceleration due to gravity.
- Therefore, the frictional force is given by:
- The retarding force acting on the car is given by:
- Where a is the acceleration of the car.
- Since the car is moving with a constant speed, a = 0.
- Therefore, the retarding force is also zero.
- For the car to stop, the frictional force should be equal to the driving force.
- Where r is the radius of curvature of the path of the car.
- Since the car is moving along a straight road, r is infinite.
- Therefore, the equation reduces to:
- This implies that the frictional force is equal to zero when the car is moving with a constant speed.
- When the brakes are applied, the frictional force acts in the opposite direction to the motion of the car.
- Therefore, the net force acting on the car is given by:
- Where the negative sign indicates that the force is acting in the opposite direction to the motion of the car.
- Therefore, the acceleration of the car is given by:
- The time taken by the car to come to a stop is given by:
- The distance travelled by the car during this time is given by:
- Substituting the given values, we
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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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer?.
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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? for NEET 2025 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for NEET 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer?.
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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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of 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) [1992]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
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