Can anyone tell me how to solve this problem?

Question

A 1600 kg car starts from rest and drives around a flat 40.0 m-diameter circular track. The forward force provided by the car's drive wheels is a constant 1000 N.What are the magnitude of the car's acceleration at t= 10.0 s?If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?( assume the coefficient of static friction to be 0.8)

Answer 1

(tangential acceleration) = (force)/(mass)
= 1000/1600 = 0.625 m/s²

(tangential velocity after 10 s) = (tangential acceleration)(time)
= (0.625)(10) = 6.25 m/s

(centripetal acceleration) = (tangential velocity)²/(radius)
= (6.25)²/40 = 0.977 m/s²

At t = 10.0 sec, the tangential acceleration is 0.625 m/s² while the centripetal acceleration is 0.977 m/s².
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(centripetal force) = (mass)(centripetal acceleration) = (1600)(0.977) = 1,562 N

(friction force) = (mass)(acc. of gravity)(coef. static friction) = (1600)(9.8)(0.8) = 12,544 N

Since 12,544 N is well greater than 1,562 N, the car will start sliding out way after 10 s. Let's calculate exactly how much time does it take,

(centripetal force) = 12,544 N = (mass)(centripetal acc.)
(centripetal acceleration) = 12,544/1600 = 7.84 m/s²

(centripetal acceleration) = 7.84 m/s² = (tangential velocity)²/(radius)
(tangential velocity) = sq. root of (7.84 x 40) = 17.7 m/s

(time) = (tangential velocity)/(tangential acceleration) = 17.7/0.625 = 28 s

It takes 28 seconds for the car to start sliding out of the circle.