A car drives down a road in such a way that its velocity (in m/sec) at time t (in sec) is v(t) = t¹/2 + 1. Find the car's average velocity (in m/sec) between t 5 and t = 7. Answer=
A car drives down a road in such a way that its velocity (in m/sec) at time t (in sec) is v(t) = t¹/2 + 1. Find the car's average velocity (in m/sec) between t 5 and t = 7. Answer=
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:
A car drives down a road in such a way that its velocity (in m/sec) at time \( t \) (in sec) is given by:
\[ v(t) = t^{1/2} + 1 \]
Find the car's average velocity (in m/sec) between \( t = 5 \) and \( t = 7 \).
### Solution:
The average velocity \( \bar{v} \) over a time interval \([t_1, t_2]\) is given by the formula:
\[ \bar{v} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt \]
Given \( t_1 = 5 \) and \( t_2 = 7 \), we need to calculate:
\[ \bar{v} = \frac{1}{7 - 5} \int_{5}^{7} (t^{1/2} + 1) \, dt \]
This can be split into two integrals:
\[ \bar{v} = \frac{1}{2} \left( \int_{5}^{7} t^{1/2} \, dt + \int_{5}^{7} 1 \, dt \right) \]
First, we calculate each integral separately:
1. **Integral of \( t^{1/2} \):**
\[ \int_{5}^{7} t^{1/2} \, dt = \int_{5}^{7} \sqrt{t} \, dt \]
Using the power rule for integration:
\[ = \left. \frac{2}{3} t^{3/2} \right|_{5}^{7} \]
Evaluating from \( 5 \) to \( 7 \):
\[ = \frac{2}{3} \left( 7^{3/2} - 5^{3/2} \right) \]
2. **Integral of \( 1 \):**
\[ \int_{5}^{7} 1 \, dt = [t]_{5}^{7} = 7 - 5 = 2 \]
Putting it all together:
\[ \bar{v} = \frac{1}{2} \left( \frac{2}{3} \left( 7^{3/2}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffb2133c9-e1e5-4d56-9c72-044227328930%2F1cc492fb-54f1-48b4-9c00-353f09ac58e3%2Fyq5brx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A car drives down a road in such a way that its velocity (in m/sec) at time \( t \) (in sec) is given by:
\[ v(t) = t^{1/2} + 1 \]
Find the car's average velocity (in m/sec) between \( t = 5 \) and \( t = 7 \).
### Solution:
The average velocity \( \bar{v} \) over a time interval \([t_1, t_2]\) is given by the formula:
\[ \bar{v} = \frac{1}{t_2 - t_1} \int_{t_1}^{t_2} v(t) \, dt \]
Given \( t_1 = 5 \) and \( t_2 = 7 \), we need to calculate:
\[ \bar{v} = \frac{1}{7 - 5} \int_{5}^{7} (t^{1/2} + 1) \, dt \]
This can be split into two integrals:
\[ \bar{v} = \frac{1}{2} \left( \int_{5}^{7} t^{1/2} \, dt + \int_{5}^{7} 1 \, dt \right) \]
First, we calculate each integral separately:
1. **Integral of \( t^{1/2} \):**
\[ \int_{5}^{7} t^{1/2} \, dt = \int_{5}^{7} \sqrt{t} \, dt \]
Using the power rule for integration:
\[ = \left. \frac{2}{3} t^{3/2} \right|_{5}^{7} \]
Evaluating from \( 5 \) to \( 7 \):
\[ = \frac{2}{3} \left( 7^{3/2} - 5^{3/2} \right) \]
2. **Integral of \( 1 \):**
\[ \int_{5}^{7} 1 \, dt = [t]_{5}^{7} = 7 - 5 = 2 \]
Putting it all together:
\[ \bar{v} = \frac{1}{2} \left( \frac{2}{3} \left( 7^{3/2}
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