Question
PROBLEM 3:A race car accelerates uniformly from a speed of 40.0
m/s to a speed of 60.0 m/s in 5.00 s while traveling counterclockwise
around a circular track of radius 4.00 x 10^2 m. When the car reaches a
speed of 50.0 m/s, calculate (a) the magnitude of the car’s centripetal
acceleration, (b) the angular velocity, (c) the magnitude of the tangential
acceleration, and (d) the magnitude of the total acceleration.
QUESTION 3: If the force causing the centripetal acceleration
suddenly vanished, would the car (a) slide away along a radius,
(b) proceed along a line tangent to the circular motion, or (c) proceed at
an angle intermediate between the tangent and radius?
EXERCISE 3: Suppose the race car now slows down uniformly from
60.0 m/s to 30.0 m/s in 4.50 s to avoid an accident, while still traversing a
circular path 4.00 x 10^2 m in radius. Calculate the car’s (a) centripetal
acceleration, (b) angular velocity, (c) tangential acceleration, and (d) total
acceleration when the speed is 40.0 m/s
Answers
Answered by
GPT-4o mini
To solve the given problems, we will use the relevant equations for centripetal acceleration, angular velocity, and tangential acceleration.
### Problem 3:
#### Given Data:
- Initial velocity, \( v_i = 40.0 \, \text{m/s} \)
- Final velocity, \( v_f = 60.0 \, \text{m/s} \)
- Time, \( t = 5.00 \, \text{s} \)
- Radius of the circular track, \( r = 4.00 \times 10^2 \, \text{m} = 400.0 \, \text{m} \)
#### (a) Calculate the magnitude of the centripetal acceleration when speed is 50.0 m/s:
**Formula:**
\[
a_c = \frac{v^2}{r}
\]
**Calculation:**
\[
a_c = \frac{(50.0 \, \text{m/s})^2}{400.0 \, \text{m}} = \frac{2500}{400} = 6.25 \, \text{m/s}^2
\]
#### (b) Calculate the angular velocity when speed is 50.0 m/s:
**Formula:**
\[
\omega = \frac{v}{r}
\]
**Calculation:**
\[
\omega = \frac{50.0 \, \text{m/s}}{400.0 \, \text{m}} = 0.125 \, \text{rad/s}
\]
#### (c) Calculate the magnitude of the tangential acceleration:
**Finding the tangential acceleration:**
To find the tangential acceleration, we can use the change in speed.
**Formula for tangential acceleration:**
\[
a_t = \frac{\Delta v}{t}
\]
**Where:**
- \( \Delta v = v_f - v_i \)
- For the speed of 50.0 m/s, it is the change from 40.0 m/s to 50.0 m/s (to avoid confusion, we can just calculate directly since speed is increasing uniformly during the first 5 seconds).
**Calculation:**
\[
a_t = \frac{v_f - v_i}{5.00 \, \text{s}} = \frac{50.0 \, \text{m/s} - 40.0 \, \text{m/s}}{5.00 \, \text{s}} = \frac{10.0 \, \text{m/s}}{5.00 \, \text{s}} = 2.0 \, \text{m/s}^2
\]
#### (d) Calculate the total acceleration:
**Total acceleration:**
The total acceleration \( a_{\text{total}} \) is the vector sum of the centripetal acceleration \( a_c \) and the tangential acceleration \( a_t \).
**Calculation:**
\[
a_{\text{total}} = \sqrt{a_c^2 + a_t^2}
\]
\[
= \sqrt{(6.25 \, \text{m/s}^2)^2 + (2.0 \, \text{m/s}^2)^2}
\]
\[
= \sqrt{39.0625 + 4}
\]
\[
= \sqrt{43.0625} \approx 6.6 \, \text{m/s}^2
\]
### Question 3:
If the force causing the centripetal acceleration suddenly vanished, the car would (b) proceed along a line tangent to the circular motion. This is consistent with Newton's first law of motion, as the object will continue in the direction of its velocity at that moment when the centripetal force is no longer acting.
### Exercise 3:
Now, we’re considering the car slowing down uniformly from 60.0 m/s to 30.0 m/s in 4.50 s while still traversing a circular path of the same radius.
#### Given Data:
- Initial velocity, \( v_i = 60.0 \, \text{m/s} \)
- Final velocity, \( v_f = 30.0 \, \text{m/s} \)
- Time, \( t = 4.50 \, \text{s} \)
#### (a) Calculate the centripetal acceleration when speed is 40.0 m/s:
**Calculation:**
\[
a_c = \frac{(40.0 \, \text{m/s})^2}{400.0 \, \text{m}} = \frac{1600}{400} = 4.00 \, \text{m/s}^2
\]
#### (b) Calculate the angular velocity when speed is 40.0 m/s:
**Calculation:**
\[
\omega = \frac{40.0 \, \text{m/s}}{400.0 \, \text{m}} = 0.1 \, \text{rad/s}
\]
#### (c) Calculate the tangential acceleration when speed is 40.0 m/s:
**Calculation of tangential acceleration:**
\[
a_t = \frac{30.0 \, \text{m/s} - 60.0 \, \text{m/s}}{4.50 \, \text{s}} = \frac{-30.0 \, \text{m/s}}{4.50 \, \text{s}} \approx -6.67 \, \text{m/s}^2
\]
(Here \( a_t \) is negative because the car is slowing down)
#### (d) Calculate the total acceleration when speed is 40.0 m/s:
**Calculation:**
\[
a_{\text{total}} = \sqrt{a_c^2 + a_t^2}
\]
\[
= \sqrt{(4.00 \, \text{m/s}^2)^2 + (-6.67 \, \text{m/s}^2)^2}
\]
\[
= \sqrt{16 + 44.4889} = \sqrt{60.4889} \approx 7.79 \, \text{m/s}^2
\]
### Final Results:
**For Problem 3:**
- (a) Centripetal acceleration: 6.25 m/s²
- (b) Angular velocity: 0.125 rad/s
- (c) Tangential acceleration: 2.0 m/s²
- (d) Total acceleration: 6.6 m/s²
**For Exercise 3:**
- (a) Centripetal acceleration: 4.00 m/s²
- (b) Angular velocity: 0.1 rad/s
- (c) Tangential acceleration: -6.67 m/s²
- (d) Total acceleration: 7.79 m/s²
### Problem 3:
#### Given Data:
- Initial velocity, \( v_i = 40.0 \, \text{m/s} \)
- Final velocity, \( v_f = 60.0 \, \text{m/s} \)
- Time, \( t = 5.00 \, \text{s} \)
- Radius of the circular track, \( r = 4.00 \times 10^2 \, \text{m} = 400.0 \, \text{m} \)
#### (a) Calculate the magnitude of the centripetal acceleration when speed is 50.0 m/s:
**Formula:**
\[
a_c = \frac{v^2}{r}
\]
**Calculation:**
\[
a_c = \frac{(50.0 \, \text{m/s})^2}{400.0 \, \text{m}} = \frac{2500}{400} = 6.25 \, \text{m/s}^2
\]
#### (b) Calculate the angular velocity when speed is 50.0 m/s:
**Formula:**
\[
\omega = \frac{v}{r}
\]
**Calculation:**
\[
\omega = \frac{50.0 \, \text{m/s}}{400.0 \, \text{m}} = 0.125 \, \text{rad/s}
\]
#### (c) Calculate the magnitude of the tangential acceleration:
**Finding the tangential acceleration:**
To find the tangential acceleration, we can use the change in speed.
**Formula for tangential acceleration:**
\[
a_t = \frac{\Delta v}{t}
\]
**Where:**
- \( \Delta v = v_f - v_i \)
- For the speed of 50.0 m/s, it is the change from 40.0 m/s to 50.0 m/s (to avoid confusion, we can just calculate directly since speed is increasing uniformly during the first 5 seconds).
**Calculation:**
\[
a_t = \frac{v_f - v_i}{5.00 \, \text{s}} = \frac{50.0 \, \text{m/s} - 40.0 \, \text{m/s}}{5.00 \, \text{s}} = \frac{10.0 \, \text{m/s}}{5.00 \, \text{s}} = 2.0 \, \text{m/s}^2
\]
#### (d) Calculate the total acceleration:
**Total acceleration:**
The total acceleration \( a_{\text{total}} \) is the vector sum of the centripetal acceleration \( a_c \) and the tangential acceleration \( a_t \).
**Calculation:**
\[
a_{\text{total}} = \sqrt{a_c^2 + a_t^2}
\]
\[
= \sqrt{(6.25 \, \text{m/s}^2)^2 + (2.0 \, \text{m/s}^2)^2}
\]
\[
= \sqrt{39.0625 + 4}
\]
\[
= \sqrt{43.0625} \approx 6.6 \, \text{m/s}^2
\]
### Question 3:
If the force causing the centripetal acceleration suddenly vanished, the car would (b) proceed along a line tangent to the circular motion. This is consistent with Newton's first law of motion, as the object will continue in the direction of its velocity at that moment when the centripetal force is no longer acting.
### Exercise 3:
Now, we’re considering the car slowing down uniformly from 60.0 m/s to 30.0 m/s in 4.50 s while still traversing a circular path of the same radius.
#### Given Data:
- Initial velocity, \( v_i = 60.0 \, \text{m/s} \)
- Final velocity, \( v_f = 30.0 \, \text{m/s} \)
- Time, \( t = 4.50 \, \text{s} \)
#### (a) Calculate the centripetal acceleration when speed is 40.0 m/s:
**Calculation:**
\[
a_c = \frac{(40.0 \, \text{m/s})^2}{400.0 \, \text{m}} = \frac{1600}{400} = 4.00 \, \text{m/s}^2
\]
#### (b) Calculate the angular velocity when speed is 40.0 m/s:
**Calculation:**
\[
\omega = \frac{40.0 \, \text{m/s}}{400.0 \, \text{m}} = 0.1 \, \text{rad/s}
\]
#### (c) Calculate the tangential acceleration when speed is 40.0 m/s:
**Calculation of tangential acceleration:**
\[
a_t = \frac{30.0 \, \text{m/s} - 60.0 \, \text{m/s}}{4.50 \, \text{s}} = \frac{-30.0 \, \text{m/s}}{4.50 \, \text{s}} \approx -6.67 \, \text{m/s}^2
\]
(Here \( a_t \) is negative because the car is slowing down)
#### (d) Calculate the total acceleration when speed is 40.0 m/s:
**Calculation:**
\[
a_{\text{total}} = \sqrt{a_c^2 + a_t^2}
\]
\[
= \sqrt{(4.00 \, \text{m/s}^2)^2 + (-6.67 \, \text{m/s}^2)^2}
\]
\[
= \sqrt{16 + 44.4889} = \sqrt{60.4889} \approx 7.79 \, \text{m/s}^2
\]
### Final Results:
**For Problem 3:**
- (a) Centripetal acceleration: 6.25 m/s²
- (b) Angular velocity: 0.125 rad/s
- (c) Tangential acceleration: 2.0 m/s²
- (d) Total acceleration: 6.6 m/s²
**For Exercise 3:**
- (a) Centripetal acceleration: 4.00 m/s²
- (b) Angular velocity: 0.1 rad/s
- (c) Tangential acceleration: -6.67 m/s²
- (d) Total acceleration: 7.79 m/s²