To solve this problem, we will break it into parts and use the relationships between linear and angular motion:

1. Angular Acceleration (α):

We are given:

• Linear deceleration ($a_t$) = $7.20 \, \text{m/s}^2$
• Radius of the tires ($r$) = $0.300 \, \text{m}$

The angular acceleration ($\alpha$) is related to the linear acceleration by: $\alpha = \frac{a_t}{r}$ Substitute the values: $\alpha = \frac{7.20}{0.300} = 24.0 \, \text{rad/s}^2$

Since the deceleration is in the opposite direction of the initial rotation, the angular acceleration will have a negative sign: $\alpha = -24.0 \, \text{rad/s}^2$

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2. Number of Revolutions (θ in revolutions):

The relationship between angular displacement ($\theta$), initial angular velocity ($\omega_0$), angular acceleration ($\alpha$), and time ($t$) is: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$

But first, we need to find the time it takes to stop.

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3. Time to Stop (t):

The angular velocity decreases uniformly until it reaches zero. The time to stop is given by: $t = \frac{\omega_0}{|\alpha|}$ Substitute the values: $t = \frac{97}{24.0} = 4.04 \, \text{seconds}$

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4. Angular Displacement ($\theta$):

The total angular displacement can be calculated using the equation: $\theta = \omega_0 t + \frac{1}{2} \alpha t^2$ Substitute the known values ($\omega_0 = 97 \, \text{rad/s}$, $t = 4.04 \, \text{s}$, and $\alpha = -24.0 \, \text{rad/s}^2$): $\theta = (97)(4.04) + \frac{1}{2}(-24.0)(4.04)^2$ First term: $97 \times 4.04 = 391.88$ Second term: $\frac{1}{2}(-24.0)(4.04)^2 = -195.84$ $\theta = 391.88 - 195.84 = 196.04 \, \text{radians}$

Convert to revolutions: $\text{Revolutions} = \frac{\theta}{2\pi} = \frac{196.04}{2\pi} \approx 31.2 \, \text{revolutions}$

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5. Distance Traveled (d):

The distance traveled by the car is related to the angular displacement by: $d = r \theta$ Substitute the values ($r = 0.300 \, \text{m}$, $\theta = 196.04 \, \text{rad}$): $d = (0.300)(196.04) \approx 58.81 \, \text{m}$

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Final Results:

1. Angular acceleration: $-24.0 \, \text{rad/s}^2$
2. Time to stop: $4.04 \, \text{seconds}$
3. Number of revolutions: $31.2 \, \text{revolutions}$
4. Distance traveled: $58.81 \, \text{m}$

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Let me know if you would like details on any of the steps or have further questions. Here are 5 questions related to this problem for you:

1. How is angular acceleration connected to linear acceleration in general?
2. Why is the angular displacement in radians converted to revolutions?
3. How would the problem change if the tires slipped on the pavement?
4. What factors influence the distance traveled during deceleration?
5. How could this problem be solved graphically instead of mathematically?

Tip: Always pay attention to unit consistency, especially when converting between radians and revolutions.