The direction of the car’s average acceleration between these two points.

Question

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Answer 1

Question

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

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Answer 2

Question

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

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Answer 3

Question

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

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

Answer 4

Question

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

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

Answer 5

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

Answer 6

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.

Answer 7

Chapter 4, Problem 1OQ

To determine

The direction of the car’s average acceleration between these two points.

Answer 8

Expert Solution & Answer

Answer to Problem 1OQ

Option (e) The resultant vector of average acceleration is directed towards South-West.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

As the car moves from point 1 to point 2, its speed doubles. Consider that speed at point 1 is v and speed at point 2 is 2v.

Write the expression for velocity vector at point 1 in xy plane as.

v→1=vj^

Here, v→1 is velocity vector at point 1 and v is the velocity of car.

Write the expression for velocity vector at point 2 in xy plane as.

v→2=2v(−i^)

Here, v→2 is the velocity vector at point 2.

Write the expression for average acceleration as

a→avg=v→2−v→1Δt (I)

Conclusion:

Substitute 2v(−i^) for v→2 and vj^ for v→1 in equation (I).

a→avg=2v(−i^)−vj^Δt=−2vi^−vj^Δt

Draw the figure for the average acceleration for the above equation as.

Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term, Chapter 4, Problem 1OQ

Thus option (e) is correct.

As per the above figure, the direction of arrow is towards North-East. The direction does not match with the resultant obtained. Thus option (a) is incorrect.

The direction of the average acceleration is towards North-West. The direction does not match with the resultant obtained. Thus option (b) is incorrect.

The direction of the average acceleration is towards North-West at some angle less that in option (b). This does not match with the resultant obtained. Thus, option (c) is incorrect.

The direction of average acceleration is towards west. This is not the resultant vector. Thus option (d) is incorrect.