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
To find the time at which the magnitude of the rock's acceleration is equal to g, we need to determine the acceleration of the rock and set it equal to the acceleration due to gravity (g).
First, let's find the acceleration of the rock. We know the tangential deceleration of the rock when the brakes are applied, which is given as 1.40 m/s^2. The tangential acceleration is the rate of change of the tangential speed, which is given by the formula:
a_t = Δv / Δt,
where a_t is the tangential acceleration, Δv is the change in tangential speed, and Δt is the change in time.
In this case, the rock's tangential acceleration is equal to the tangential deceleration, so we have:
1.40 m/s^2 = Δv / Δt.
Next, we need to relate the tangential acceleration to the magnitude of acceleration. The magnitude of acceleration is given by the formula:
a = √(a_t^2 + a_n^2),
where a is the magnitude of acceleration, a_t is the tangential acceleration, and a_n is the normal acceleration. In this case, we can assume the normal acceleration to be zero since there is no vertical motion.
Therefore, we have:
a = a_t.
Now, we can set the magnitude of acceleration equal to g:
g = a.
Substituting the expression for tangential acceleration, we have:
g = 1.40 m/s^2.
Now that we have the value for g, we can find the time at which the magnitude of the rock's acceleration is equal to g by rearranging the equation:
1.40 m/s^2 = g.
Solving for Δt, we get:
Δt = Δv / 1.40 m/s^2.
Since the tangential speed is given as 3.00 m/s, we have:
Δt = 3.00 m/s / 1.40 m/s^2.
Evaluating this expression, we find:
Δt ≈ 2.14 s.
Therefore, at approximately 2.14 seconds, the magnitude of the rock's acceleration is equal to g.