You are driving a car away from home. Your velocity (miles per hour) t hours after noon is given by v(t) = -5t +45t-147t + 180t. At noon you were 160 miles from home. At 2:45pm you were driving at a rate of miles per hour. (Round answer to nearest tenth.)
You are driving a car away from home. Your velocity (miles per hour) t hours after noon is given by v(t) = -5t +45t-147t + 180t. At noon you were 160 miles from home. At 2:45pm you were driving at a rate of miles per hour. (Round answer to nearest tenth.)
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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![**Problem Statement:**
You are driving a car away from home. Your velocity (miles per hour) \( t \) hours after noon is given by:
\[ v(t) = -5t^4 + 45t^3 - 147t^2 + 180t. \]
At noon you were 160 miles from home.
At 2:45 pm you were driving at a rate of \(\square\) miles per hour. (Round answer to nearest tenth.)
---
**Instructions:**
1. Identify the time elapsed from noon to 2:45 pm:
- From noon to 2:45 pm is 2 hours and 45 minutes.
- Convert 45 minutes to hours: \( \frac{45}{60} = 0.75 \) hours.
- Therefore, \( t = 2.75 \) hours.
2. Substitute \( t = 2.75 \) into the velocity function \( v(t) \) to find the rate of velocity at 2:45 pm:
\[ v(2.75) = -5(2.75)^4 + 45(2.75)^3 - 147(2.75)^2 + 180(2.75). \]
3. Perform the calculations to determine the velocity at 2:45 pm:
- Calculate the individual terms:
\[ (2.75)^2 = 7.5625, \]
\[ (2.75)^3 = 20.796875, \]
\[ (2.75)^4 = 57.19140625. \]
- Substitute these values into the equation:
\[ v(2.75) = -5(57.19140625) + 45(20.796875) - 147(7.5625) + 180(2.75). \]
- Simplify each term:
\[ -5 \times 57.19140625 = -285.95703125, \]
\[ 45 \times 20.796875 = 935.859375, \]
\[ -147 \times 7.5625 = -1111.6875, \]
\[ 180 \times 2.75 = 495. \]
- Combine the results:
\[ v(2.75](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4b0b4e2-88c2-4896-b905-dcd3cba1ae99%2Fad0d3dfd-135d-414b-ab4f-300cb171cf82%2Fv0net65_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
You are driving a car away from home. Your velocity (miles per hour) \( t \) hours after noon is given by:
\[ v(t) = -5t^4 + 45t^3 - 147t^2 + 180t. \]
At noon you were 160 miles from home.
At 2:45 pm you were driving at a rate of \(\square\) miles per hour. (Round answer to nearest tenth.)
---
**Instructions:**
1. Identify the time elapsed from noon to 2:45 pm:
- From noon to 2:45 pm is 2 hours and 45 minutes.
- Convert 45 minutes to hours: \( \frac{45}{60} = 0.75 \) hours.
- Therefore, \( t = 2.75 \) hours.
2. Substitute \( t = 2.75 \) into the velocity function \( v(t) \) to find the rate of velocity at 2:45 pm:
\[ v(2.75) = -5(2.75)^4 + 45(2.75)^3 - 147(2.75)^2 + 180(2.75). \]
3. Perform the calculations to determine the velocity at 2:45 pm:
- Calculate the individual terms:
\[ (2.75)^2 = 7.5625, \]
\[ (2.75)^3 = 20.796875, \]
\[ (2.75)^4 = 57.19140625. \]
- Substitute these values into the equation:
\[ v(2.75) = -5(57.19140625) + 45(20.796875) - 147(7.5625) + 180(2.75). \]
- Simplify each term:
\[ -5 \times 57.19140625 = -285.95703125, \]
\[ 45 \times 20.796875 = 935.859375, \]
\[ -147 \times 7.5625 = -1111.6875, \]
\[ 180 \times 2.75 = 495. \]
- Combine the results:
\[ v(2.75
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