Answer :


The maximum force of friction is developed between the surface of road and the car’s tyres when hard breaks are applied.


So, maximum frictional force = μR


From the free body diagram,


R mg cos θ = 0
R = mg cos θ (i)
and
μR + mamg sin θ = 0 (ii)
μ mg cos θ + mamg sin θ = 0


μg cos θ + a -10(12) =0


a=5- {1-(2)} ×10(/2)


a = -2.5 m/s2


When brakes are applied, car will deaccelerate by 2.5 m/s2


Distance s= 12.8 m
initial velocity u = 6 m/s
Velocity at the end of incline


ν=



v=10 m/s=36 km/h


Hence, the harder the driver applies the brakes, the lower will be the velocity of the car when it reaches the ground, i.e. at 36 km/h


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses
RELATED QUESTIONS :

A rectangular boxPhysics - Exemplar

A helicopter of mPhysics - Exemplar

There are four foPhysics - Exemplar

When a body <spanPhysics - Exemplar

Block <span lang=Physics - Exemplar

A block <span lanPhysics - Exemplar

Two <span lang="EPhysics - Exemplar