Final answer:
a. each tire makes approximately 157.75 revolutions before the car comes to a stop.
b. the angular speed of the wheels when the car has traveled half the total distance is approximately \(81.59 \, \text{rad/s}\).
Step-by-step explanation:
Let's start by solving part (a) to find the number of revolutions each tire makes before the car comes to a stop.
(a) To find the number of revolutions (\(N\)), we can use the following kinematic equation:
\[v_f^2 = v_i^2 + 2a d\]
Where:
- \(v_f\) is the final velocity (which is 0 since the car comes to a stop),
- \(v_i\) is the initial velocity,
- \(a\) is the acceleration,
- \(d\) is the displacement.
Rearranging the equation to solve for \(d\):
\[d = \frac{{v_f^2 - v_i^2}}{{2a}}\]
Now, we can use the relationship between linear distance (displacement) and angular distance (number of revolutions) for a wheel:
\[d = N \cdot 2\pi r\]
Where:
- \(N\) is the number of revolutions,
- \(r\) is the radius of the tire.
Substitute the expression for \(d\) into this equation and solve for \(N\):
\[N = \frac{{v_f^2 - v_i^2}}{{2a \cdot 2\pi r}}\]
Now plug in the values: \(v_i = 25.7 \, \text{m/s}\), \(a = -2.00 \, \text{m/s}^2\) (negative because it's deceleration), \(r = 0.315 \, \text{m}\), and \(v_f = 0\).
(b) To find the angular speed of the wheels when the car has traveled half the total distance, we can use the formula:
\[ \text{Angular Speed} (\omega) = \frac{v}{r} \]
Let's solve both parts.
**Part (a):**
\[ N = \frac{{0^2 - (25.7 \, \text{m/s})^2}}{{2 \cdot (-2.00 \, \text{m/s}^2) \cdot 2\pi \cdot 0.315 \, \text{m}}}\]
Calculate \(N\) using the given values.
**Part (b):**
Once you have the value of \(N\), you can find the distance traveled (\(d\)) using the relationship \(d = N \cdot 2\pi r\). Then, find the total distance traveled (\(D\)) and determine the distance at which half the total distance is covered. Finally, use this distance to find the angular speed using \(\omega = \frac{v}{r}\).
I apologize for the incomplete response. Let's proceed with the calculations.
**Part (a):**
\[ N = \frac{{0^2 - (25.7 \, \text{m/s})^2}}{{2 \cdot (-2.00 \, \text{m/s}^2) \cdot 2\pi \cdot 0.315 \, \text{m}}}\]
\[ N \approx \frac{{-660.49}}{{-4.188}} \]
\[ N \approx 157.75 \]
So, each tire makes approximately 157.75 revolutions before the car comes to a stop.
**Part (b):**
Now, let's find the distance traveled (\(d\)) using the relationship \(d = N \cdot 2\pi r\):
\[ d = 157.75 \cdot 2\pi \cdot 0.315 \, \text{m} \]
\[ d \approx 313.55 \, \text{m} \]
Half the total distance (\(D\)) is \(D/2 = 313.55/2 = 156.78 \, \text{m}\).
Now, use the angular speed formula \(\omega = \frac{v}{r}\) to find the angular speed when the car has traveled half the total distance:
\[ \omega = \frac{25.7 \, \text{m/s}}{0.315 \, \text{m}} \]
\[ \omega \approx 81.59 \, \text{rad/s} \]
So, the angular speed of the wheels when the car has traveled half the total distance is approximately \(81.59 \, \text{rad/s}\).