A car traveling east at 35.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.80 s. (a) What is the magnitude and direction of the car's acceleration as it slows down? magnitude |m/s² direction --Select--- v (b) How far does the car travel in the 3.80-s time period?
A car traveling east at 35.0 m/s passes a trooper hiding at the roadside. The driver uniformly reduces his speed to 25.0 m/s in 3.80 s. (a) What is the magnitude and direction of the car's acceleration as it slows down? magnitude |m/s² direction --Select--- v (b) How far does the car travel in the 3.80-s time period?
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![### Example Problem: Deceleration of a Car
A car traveling east at **35.0 m/s** passes a trooper hiding at the roadside. The driver uniformly reduces his speed to **25.0 m/s** in **3.80 s**.
#### (a) What is the magnitude and direction of the car's acceleration as it slows down?
- **Magnitude:** $\_\_\_\_\_$ m/s²
- **Direction:** $\_\_\_\_$ (select from the dropdown)
#### (b) How far does the car travel in the **3.80-s** time period?
- $\_\_\_\_$ m
**Explanation:**
1. **Acceleration Calculation (a):**
To determine the magnitude and direction of the car's acceleration, we will use the formula for acceleration:
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval.
The change in velocity \( \Delta v \):
\[
\Delta v = v_{\text{final}} - v_{\text{initial}} = 25.0 \, \text{m/s} - 35.0 \, \text{m/s} = -10.0 \, \text{m/s}
\]
The time interval \( \Delta t \) is given as 3.80 s.
Thus:
\[
a = \frac{-10.0 \, \text{m/s}}{3.80 \, \text{s}} \approx -2.63 \, \text{m/s}^2
\]
The negative sign indicates the direction of acceleration is opposite to the direction of motion, which means it is slowing down (deceleration).
2. **Distance Calculation (b):**
To find how far the car travels during this period, we can use the formula for distance under uniform acceleration:
\[
s = v_{\text{initial}} \cdot t + \frac{1}{2} a t^2
\]
Plugging in the values:
\[
s = (35.0 \, \text{m/s}) \cdot (3.80 \, \text{s}) + \frac{1}{2} (-2.63 \,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6b165613-2027-4238-9c7e-a40fd74e53eb%2F5305788a-cb74-4d10-a9f5-1c2ea4db0f8c%2Fblciupi.png&w=3840&q=75)
Transcribed Image Text:### Example Problem: Deceleration of a Car
A car traveling east at **35.0 m/s** passes a trooper hiding at the roadside. The driver uniformly reduces his speed to **25.0 m/s** in **3.80 s**.
#### (a) What is the magnitude and direction of the car's acceleration as it slows down?
- **Magnitude:** $\_\_\_\_\_$ m/s²
- **Direction:** $\_\_\_\_$ (select from the dropdown)
#### (b) How far does the car travel in the **3.80-s** time period?
- $\_\_\_\_$ m
**Explanation:**
1. **Acceleration Calculation (a):**
To determine the magnitude and direction of the car's acceleration, we will use the formula for acceleration:
\[
a = \frac{\Delta v}{\Delta t}
\]
where \( \Delta v \) is the change in velocity and \( \Delta t \) is the time interval.
The change in velocity \( \Delta v \):
\[
\Delta v = v_{\text{final}} - v_{\text{initial}} = 25.0 \, \text{m/s} - 35.0 \, \text{m/s} = -10.0 \, \text{m/s}
\]
The time interval \( \Delta t \) is given as 3.80 s.
Thus:
\[
a = \frac{-10.0 \, \text{m/s}}{3.80 \, \text{s}} \approx -2.63 \, \text{m/s}^2
\]
The negative sign indicates the direction of acceleration is opposite to the direction of motion, which means it is slowing down (deceleration).
2. **Distance Calculation (b):**
To find how far the car travels during this period, we can use the formula for distance under uniform acceleration:
\[
s = v_{\text{initial}} \cdot t + \frac{1}{2} a t^2
\]
Plugging in the values:
\[
s = (35.0 \, \text{m/s}) \cdot (3.80 \, \text{s}) + \frac{1}{2} (-2.63 \,
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