Posted by **skywalker** on Saturday, November 27, 2010 at 5:50pm.

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?

- physics -
**drwls**, Saturday, November 27, 2010 at 6:12pm
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

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