Stopping Distance A car is traveling at a rate of 50 feet per second when the brakes are applied. The position function for the car is given by s = -6.25t2 + 50t, where s is measured in feet and t is measured in seconds. Use this function to complete the table showing the position, velocity, and acceleration for each given value of t. t 3 4 ds dt d?s What can you conclude? It takes seconds for the car to stop, at which time it has traveled ft.
Stopping Distance A car is traveling at a rate of 50 feet per second when the brakes are applied. The position function for the car is given by s = -6.25t2 + 50t, where s is measured in feet and t is measured in seconds. Use this function to complete the table showing the position, velocity, and acceleration for each given value of t. t 3 4 ds dt d?s What can you conclude? It takes seconds for the car to stop, at which time it has traveled ft.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![**Stopping Distance**
A car is traveling at a rate of 50 feet per second when the brakes are applied. The position function for the car is given by \( s = -6.25t^2 + 50t \), where \( s \) is measured in feet and \( t \) is measured in seconds. Use this function to complete the table showing the position, velocity, and acceleration for each given value of \( t \).
| \( t \) | 0 | 1 | 2 | 3 | 4 |
|---------|---|---|---|---|---|
| \( s \) | | | | | |
| \( \frac{ds}{dt} \) | | | | | |
| \( \frac{d^2s}{dt^2} \) | | | | | |
**What can you conclude?**
It takes [ ] seconds for the car to stop, at which time it has traveled [ ] ft.
**Explanation**
To solve this, we need to calculate:
1. **Position (\( s \))**: Use the function \( s = -6.25t^2 + 50t \) to find the position at each time \( t \).
2. **Velocity (\( \frac{ds}{dt} \))**: Differentiate the position function to find the velocity. The derivative is \( \frac{ds}{dt} = -12.5t + 50 \).
3. **Acceleration (\( \frac{d^2s}{dt^2} \))**: Differentiate the velocity function to find the acceleration. The derivative is \( \frac{d^2s}{dt^2} = -12.5 \).
Use these equations to fill in the table and answer the questions about the time taken to stop and the total distance traveled.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F9b42a4e3-d8c3-4e59-af97-f04240a64541%2F23f5796f-4a9a-4db8-99b5-eb9e01ba823c%2Fbj7ptx_processed.png&w=3840&q=75)
Transcribed Image Text:**Stopping Distance**
A car is traveling at a rate of 50 feet per second when the brakes are applied. The position function for the car is given by \( s = -6.25t^2 + 50t \), where \( s \) is measured in feet and \( t \) is measured in seconds. Use this function to complete the table showing the position, velocity, and acceleration for each given value of \( t \).
| \( t \) | 0 | 1 | 2 | 3 | 4 |
|---------|---|---|---|---|---|
| \( s \) | | | | | |
| \( \frac{ds}{dt} \) | | | | | |
| \( \frac{d^2s}{dt^2} \) | | | | | |
**What can you conclude?**
It takes [ ] seconds for the car to stop, at which time it has traveled [ ] ft.
**Explanation**
To solve this, we need to calculate:
1. **Position (\( s \))**: Use the function \( s = -6.25t^2 + 50t \) to find the position at each time \( t \).
2. **Velocity (\( \frac{ds}{dt} \))**: Differentiate the position function to find the velocity. The derivative is \( \frac{ds}{dt} = -12.5t + 50 \).
3. **Acceleration (\( \frac{d^2s}{dt^2} \))**: Differentiate the velocity function to find the acceleration. The derivative is \( \frac{d^2s}{dt^2} = -12.5 \).
Use these equations to fill in the table and answer the questions about the time taken to stop and the total distance traveled.
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