Circular Motion Physics Problem.. NEED HELP!?

1) A car is traveling at 18.4 m/s on a straight horizontal highway. The wheels of the car have radii of 49.0 cm. If the car speeds up with an acceleration of 1.50 m/s2 for 4.80 s, find the number of revolutions of the wheels during this period.








2)A race car starts from rest on a circular track of radius 250 m. The car's speed increases at the constant rate of 0.360 m/s2. At the point where the magnitudes of the centripetal and tangential accelerations are equal, determine


a) the speed of the race car in (m/s)


b) the distance traveled in (m)


c) the elapsed time in (s)





3)A 50.0 kg child stands at the rim of a merry-go-round of radius 2.20 m,rotating with an angular speed of 2.70 rad/s.


a) What is the child's centripetal acceleration?


b) What is the minimum force between her feet and the floor of the merry-go-round that is required to keep her in the circular path(N)?


c) What minimum coefficient of static friction is required?


is she likely to be able to stay?

Circular Motion Physics Problem.. NEED HELP!?
1. The initial velocity is 18.4 m/s and the acceleration is 1.50 m/s/s. Therefore the distance travelled in 4.80 s is 18.4 m/s * 4.80 s + 1/2(1.50m/s/s)(4.80 s)^2 which comes out to 106 m. The circumference of the wheels is .980 m * pi = 3.08 m. To find the revolutions you divide the distance by the circumference 106m/3.08m to find 34.4 revolutions.





2.





a. The tangential acceleration is a constant .360 m/s/s therefore you merely need to solve .360 m/s/s = (v^2)/250.m. For this equation v = 9.49 m/s.





b. V^2 = 2a(D) where D is change in displacement. Solving for D we get (90m*m/s/s)/(2*1.5 m/s/s) = 30 m.





c. Time is equal to sqrt(2D/a) = sqrt(60/1.5) = 6.32 s.





3.


a. The equation for centripetal acceleration is (V^2)/r. V is 2.20 m * 2.70 rad/s. Thus the equation is simply (2.70 rad/s)^2*(2.20 m) = 16.0 m/s/s.





b. To go at that velocity with this radius she must have the specified centripetal acceleration, 16.0 m/s/s at every point. Therefore the force required is 16.0m/s/s * 50.0 kg = 800. N





c. The coefficient of friction is proportional to the normal force which is the opposite of the weight. Thus we are looking that 16.0 m/s/s * m = k*9.80m/s/s*m where m is mass and k is the coefficient of friction. The masses cancel leaving 16.0 m/s/s = k*9.80 m/s/s with k being equivalent to 1.63. So uhh, that's pretty high even for rubber, so I'd say no.