Find the average velocity and instantaneous velocity of the given particle.

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Answer 1

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Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

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Answer 2

Question

100%

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

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Answer 3

Question

100%

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

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Answer 4

Question

100%

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

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Answer 5

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Answer 6

Question

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Answer 7

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

Answer 8

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

Answer 9

Chapter 2, Problem 1CRE

To determine

Find the average velocity and instantaneous velocity of the given particle.

Answer 10

Expert Solution & Answer

Answer to Problem 1CRE

Solution:

Average velocity [2, 5] = 0.954 ms^-1

Instantaneous velocity = 0.894 ms^-1

Answer 11

Expert Solution & Answer

Answer 12

Expert Solution & Answer

Answer 13

Explanation of Solution

Given:

$\begin{array}{l} s (t) = \sqrt{t^{2} + 1} \\ t \in [2, 5] \end{array}$

Formula used:

$\begin{array}{l} Average velocity = \frac{Displacement Change}{Time Change} \\ Instantaneous velocity = \frac{d (Displacement)}{d (Time)} \end{array}$

Calculation:

$Average velocity = \frac{Displacement Change}{Time Change}$

Displacement at t = 2

$\begin{array}{l} S (2) = \sqrt{2^{2} + 1} = \sqrt{5} \\ S (5) = \sqrt{5^{2} + 1} = \sqrt{26} \\ Average velocity = \frac{Displacement Change}{Time Change} \\ Average velocity = \frac{\sqrt{26} - \sqrt{5}}{5 - 2} = \frac{5.09 - 2.23}{3} = \frac{2.86}{3} = 0.954 m s^{- 1} \\ Instantaneous velocity = \frac{d (\sqrt{t^{2} + 1})}{d t} \\ Instantaneous velocity = \frac{1}{2} (^{t^{2} + 1) - \frac{1}{2}} (2 t) = \frac{2 t}{2 \sqrt{(t^{2} + 1)}} \\ Velocity (t = 2) = \frac{2 \times 2}{2 \sqrt{(2^{2} + 1)}} = \frac{4}{2 (\sqrt{5})} = \frac{2}{\sqrt{5}} = 0.894 m s^{- 1} \end{array}$

Conclusion:

To find the average velocity we can input values in the function and calculate the value. But when finding the instantaneous velocity we have to differentiate the displacement function and input the values.

Answer 14

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E

Find the average velocity and instantaneous velocity of the given particle.

Find the average velocity and instantaneous velocity of the given particle.

Answer to Problem 1CRE

Explanation of Solution

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Chapter 2 Solutions