1800 kg car starts from rest and drives around a flat 66--diameter circular trac

Question

1800 kg car starts from rest and drives around a flat 66--diameter circular track. The forward force provided by the car's drive wheels is a constant 1100N

What is the magnitude of the car's acceleration at t= 11Sec....i got this correct....1.5 m/s^2

What is the direction of the car's acceleration at t= 11s. Give the direction as an angle from the r-axis......this one answer is 24 degrees

need the answer for this......

If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle????

Explanation / Answer

the centripetal force is given as the following:
(m*v^2)/r = (mu)*m*g, where (mu) is the constant for rubber on concrete which actually equals 1.
so simplifying you would get that v is the sq root of (mu)*r*g
substituting v = sq.root (1)*(33)*(9.8)
v = 17.98 m/s
using the all famous equation vf = vi + a(tan)*(delta t) and vi = 0 we get a result of
(delta t) = vf / a(tan)
substituting, (delta t) = 17.98 / 0.611
(delta t) = 29.43 seconds.

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