Asked by skywalker
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?
At what time is the magnitude of the rock's acceleration equal to g?
Answers
Answered by
drwls
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
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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