A1500 kg car drives around a flat, 50 m diameter track, starting from rest. the drive wheels supply a small but steady 525 n force in a forward direction. the coefficient of static friction between the tires and the road is 0.90. how many revolutions of the track have been made when the car slides out?

No. of revolutions before the car slides off is 2

Solution:

As per the question:

Mass of the car, m = 1500 kg

Diameter of the track, d = 50 m

Force, F = 525 N

Coefficient of static friction,

Now,

The acceleration in the tangential direction is given by:

Here, the centripetal force is given by the friction force:

Thus

Time taken by the car is given by:

v = v' + at

v' = initial velocity = 0

Thus

Total Distance covered, d is given by:

d = 341.99 m

Distance covered in 1 revolution is the circumference of the circle, d' =

Now, the no. of revolutions is given by:

a and c

c is 100% but i am confused in a