A car initially at rest is on a floor with coefficient if friction 0.5 . Find the minimum distance in which the car can attain a speed of 72km/hr .
With proper explanation
Answers
Answered by
2
Normal reaction (N) = mg
Limiting friction = μN = μmg
Acceleration (a) = -(Limiting friction / Mass)
= -μmg / m
= -μg
[Negative sign indicates that friction is acting in the direction opposite to the motion of car]
72 kmph = 72 × 5/18 m/s = 20 m/s
Now, use equation of motion
S = -u² / (2a) ………[∵ v = 0]
= -u² / (2 × (-μg))
= u² / (2 μg)
= (20 m/s)² / (2 × 0.5 × 10 m/s²)
= 40 m
Minimum distance car can travel is 40 m
Limiting friction = μN = μmg
Acceleration (a) = -(Limiting friction / Mass)
= -μmg / m
= -μg
[Negative sign indicates that friction is acting in the direction opposite to the motion of car]
72 kmph = 72 × 5/18 m/s = 20 m/s
Now, use equation of motion
S = -u² / (2a) ………[∵ v = 0]
= -u² / (2 × (-μg))
= u² / (2 μg)
= (20 m/s)² / (2 × 0.5 × 10 m/s²)
= 40 m
Minimum distance car can travel is 40 m
Similar questions