A race car starts from rest on a circular track of radius 400 m. Its speed increases at the constant rate of 0.500 m/s2. At the point where the magnitudes of the radial acceleration is twice the tangential acceleration, determine (a) the speed of the race car, and (b) the elapsed time.
3 answers
hey wants the answer?
tangential a = 0.5 m/s^2
so
v = 0.5 t
radial a = v^2/R = 0.25 t^2/ 400 = 0.000625 t^2
at radial acc = 1.0 m/s^2 which is twice a
1.0 = 0.000625 t^2
t^2 = 1600
t = 40 seconds
v = 0.5 t = 20 m/s
so
v = 0.5 t
radial a = v^2/R = 0.25 t^2/ 400 = 0.000625 t^2
at radial acc = 1.0 m/s^2 which is twice a
1.0 = 0.000625 t^2
t^2 = 1600
t = 40 seconds
v = 0.5 t = 20 m/s
big hurry?