NEET Exam > NEET Questions > Consider a car moving along a straight horiza...
Start Learning for Free
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[1994]
- a)30 m
- b)40 m
- c)72 m
- d)20 m
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
Consider a car moving along a straight horizantal road with a speed of...
Force due to friction = kinetic energy
[Here, v = 72 km/h
Most Upvoted Answer
Consider a car moving along a straight horizantal road with a speed of...
To find the shortest distance in which the car can be stopped, we need to consider the forces acting on the car and the equations of motion.
Given data:
Speed of the car (v) = 72 km/h
Coefficient of static friction (μ) = 0.5
We need to convert the speed from km/h to m/s:
1 km/h = 1000 m/3600 s = 5/18 m/s
So, the speed of the car in m/s is:
v = 72 km/h × 5/18 m/s = 20 m/s
Now, let's analyze the forces acting on the car:
1. The force of friction (f) acts in the opposite direction to the motion and is given by the equation:
f = μN
where N is the normal force exerted by the road on the car.
2. The normal force (N) is equal to the weight of the car (mg), where m is the mass of the car and g is the acceleration due to gravity.
3. The force of friction can also be expressed as:
f = ma
where a is the acceleration of the car.
Since the car is being brought to a stop, its final velocity (vf) is 0 m/s.
Now, using the equation of motion:
vf^2 = vi^2 + 2as
where vi is the initial velocity, a is the acceleration, and s is the distance.
Substituting the given values:
0^2 = (20 m/s)^2 + 2a(s)
0 = 400 m^2/s^2 + 2as
-400 m^2/s^2 = 2as
Now, substituting the value of the force of friction (f) in terms of acceleration (a):
-400 m^2/s^2 = 2f/m × s
-400 m^2/s^2 = 2(μN)/m × s
-400 m^2/s^2 = 2(μmg)/m × s
-400 m^2/s^2 = 2μg × s
Simplifying the equation:
-400 = 2 × 0.5 × 9.8 × s
-400 = 9.8s
s = -400/9.8
s ≈ -40.81 m
The negative sign indicates that the distance is in the opposite direction to the motion. However, distance cannot be negative, so we take the magnitude of the distance:
s ≈ 40.81 m
Therefore, the shortest distance in which the car can be stopped is approximately 40.81 meters.
Given data:
Speed of the car (v) = 72 km/h
Coefficient of static friction (μ) = 0.5
We need to convert the speed from km/h to m/s:
1 km/h = 1000 m/3600 s = 5/18 m/s
So, the speed of the car in m/s is:
v = 72 km/h × 5/18 m/s = 20 m/s
Now, let's analyze the forces acting on the car:
1. The force of friction (f) acts in the opposite direction to the motion and is given by the equation:
f = μN
where N is the normal force exerted by the road on the car.
2. The normal force (N) is equal to the weight of the car (mg), where m is the mass of the car and g is the acceleration due to gravity.
3. The force of friction can also be expressed as:
f = ma
where a is the acceleration of the car.
Since the car is being brought to a stop, its final velocity (vf) is 0 m/s.
Now, using the equation of motion:
vf^2 = vi^2 + 2as
where vi is the initial velocity, a is the acceleration, and s is the distance.
Substituting the given values:
0^2 = (20 m/s)^2 + 2a(s)
0 = 400 m^2/s^2 + 2as
-400 m^2/s^2 = 2as
Now, substituting the value of the force of friction (f) in terms of acceleration (a):
-400 m^2/s^2 = 2f/m × s
-400 m^2/s^2 = 2(μN)/m × s
-400 m^2/s^2 = 2(μmg)/m × s
-400 m^2/s^2 = 2μg × s
Simplifying the equation:
-400 = 2 × 0.5 × 9.8 × s
-400 = 9.8s
s = -400/9.8
s ≈ -40.81 m
The negative sign indicates that the distance is in the opposite direction to the motion. However, distance cannot be negative, so we take the magnitude of the distance:
s ≈ 40.81 m
Therefore, the shortest distance in which the car can be stopped is approximately 40.81 meters.
|
Explore Courses for NEET exam
|
|
Top Courses for NEETView all
Question Description
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[1994]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 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[1994]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 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[1994]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 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[1994]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 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[1994]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 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[1994]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer?.
Solutions for 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[1994]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for NEET.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
Here you can find the meaning of 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[1994]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
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[1994]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 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[1994]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 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[1994]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice 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[1994]a)30 mb)40 mc)72 md)20 mCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice NEET tests.
|
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
|
© EduRev
|
Education Revolution
|
|
Signup to see your scores
go up
within 7 days!
within 7 days!
Takes less than 10 seconds to signup