physics
posted by skywalker on .
A rock stuck in the tread of a 57.0 cm diameter bicycle wheel has a tangential speed of 3.00 m/s. When the brakes are applied, the rock's tangential deceleration is 1.40 m/s^2.
At what time is the magnitude of the rock's acceleration equal to g?

The magnitude of the rock's acceleration during deceleration in is
sqrt[(a_t)^2 + (V^2/R)^2]
Set that equal to g = 9.8 m/s^2
fo get the velocity V when the acceleratikon magnitude is g.
The tangential acceleration is
a_t = 1.4 m/s^2
Therefore
9.8^2 = 1.96 + V^4/2
V^2/R = 9.7 m/s^2
Since R = 0.285 m,
V = 1.66 m/s
Solve for the time t when
1.66 = 3.00  1.40 t m/s
V = 3.00  1.4t m^2