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

Question

Want to see more full solutions like this?

Answer 1

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

Only 100% sure experts solve it correct complete solutions ok

rmine the immediate settlement for points A and B shown in figure below knowing that Aq,-200kN/m², E-20000kN/m², u=0.5, Depth of foundation (DF-0), thickness of layer below footing (H)=20m. 4m B 2m 2m A 2m + 2m 4m

sy = f(x) + + + + + + + + + X 3 4 5 7 8 9 The function of shown in the figure is continuous on the closed interval [0, 9] and differentiable on the open interval (0, 9). Which of the following points satisfies conclusions of both the Intermediate Value Theorem and the Mean Value Theorem for f on the closed interval [0, 9] ? (A A B B C D

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Question

100%

Answer 6

Question

100%

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. 5PQ 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

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

Want to see more full solutions like this?

Chapter 2 Solutions