Let's solve each part step by step.

Given Data:

• Angular acceleration of the wheels, $\alpha = 1.10 \, \text{rad/s}^2$
• Radius of the wheels, $r = 33.0 \, \text{cm} = 0.33 \, \text{m}$

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(a) Find the magnitude of the bicycle's linear acceleration.

The linear acceleration $a$ is related to the angular acceleration $\alpha$ by the formula: $a = \alpha \cdot r$ Substitute the values: $a = 1.10 \, \text{rad/s}^2 \times 0.33 \, \text{m} = 0.363 \, \text{m/s}^2$

Answer: The linear acceleration is $0.363 \, \text{m/s}^2$.

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(b) Find the angular speed of the wheels when the linear speed of the bicyclist reaches $11.2 \, \text{m/s}$.

The relationship between linear speed $v$ and angular speed $\omega$ is: $v = \omega \cdot r$ Solving for $\omega$: $\omega = \frac{v}{r}$ Substitute the values: $\omega = \frac{11.2 \, \text{m/s}}{0.33 \, \text{m}} = 33.94 \, \text{rad/s}$

Answer: The angular speed of the wheels is $33.94 \, \text{rad/s}$.

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(c) Find how many radians the wheels have turned through in that time.

First, find the time $t$ it took for the bicycle to reach a linear speed of $11.2 \, \text{m/s}$.

Using the linear acceleration $a = 0.363 \, \text{m/s}^2$: $v = a \cdot t \Rightarrow t = \frac{v}{a} = \frac{11.2 \, \text{m/s}}{0.363 \, \text{m/s}^2} = 30.86 \, \text{s}$

Now, we use the angular displacement formula for constant angular acceleration: $\theta = \omega_0 \cdot t + \frac{1}{2} \alpha t^2$ Since the bicycle starts from rest, $\omega_0 = 0$: $\theta = \frac{1}{2} \cdot \alpha \cdot t^2$ Substitute the values: $\theta = \frac{1}{2} \times 1.10 \, \text{rad/s}^2 \times (30.86 \, \text{s})^2$ $\theta = 524.04 \, \text{rad}$

Answer: The wheels have turned through $524.04 \, \text{radians}$.

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(d) Find how far the bicycle has traveled in that time.

The linear distance $s$ traveled is related to the angular displacement $\theta$ by: $s = \theta \cdot r$ Substitute the values: $s = 524.04 \, \text{rad} \times 0.33 \, \text{m} = 172.93 \, \text{m}$

Answer: The bicycle has traveled $172.93 \, \text{m}$.

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Summary of Answers:

• (a) Linear acceleration: $0.363 \, \text{m/s}^2$
• (b) Angular speed: $33.94 \, \text{rad/s}$
• (c) Angular displacement: $524.04 \, \text{rad}$
• (d) Distance traveled: $172.93 \, \text{m}$

Do you have any questions on these calculations or need further details?

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Related Questions:

1. How would the answers change if the wheel radius were different?
2. What if the angular acceleration varied over time instead of being constant?
3. How would the total time change if the linear speed was halved?
4. What is the relationship between the linear and angular quantities in rotational motion?
5. How do centripetal forces relate to the motion of a bicycle wheel?

Tip:

Always double-check units when converting between linear and angular motion to ensure consistent results.