Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
Bundle: Physics for Scientists and Engineers with Modern Physics, Loose-leaf Version, 9th + WebAssign Printed Access Card, Multi-Term
9th Edition
ISBN: 9781305932302
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 3, Problem 46P

(a)

To determine

The position vector of point A in terms of fraction f .

(a)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A in terms of fraction f is A=[(5+11f)i^+(3+9f)j^]m .

Explanation of Solution

Section 1:

To determine: The horizontal position of point A that must be at a fraction f of the destination.

Answer: The horizontal position of point A that must be at a fraction f of the destination is (5+11f)m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the x-position of point A that must be at a fraction f of the destination is,

x=x1+f(x2x1)

  • x is the position of the point A in x-direction.
  • x1 is the initial position in x-direction.
  • x2 is the final position in x-direction.
  • f is the fraction of the destination path.

Substitute 5m for x1 and 16m for x2 .

x=5m+f(16m5m)=5m+f(11m)=(5+11f)m

Section 2:

To determine: The vertical position of point A that must be at a fraction f of the destination.

Answer: The vertical position of point A that must be at a fraction f of the destination is (3+9f)m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the y-position of point A that must be at a fraction of f of the destination is,

y=y1+f(y2y1)

  • y is the position of the point A in y-direction.
  • y1 is the initial position in y-direction.
  • y2 is the final position in y-direction.
  • f is the fraction of the destination path.

Substitute 3m for y1 and 12m for y2 .

y=3m+f(13m3m)=3m+f(9m)=(3+9f)m

Section 3:

To determine: The position vector of the point A that must lie at a fraction f on the destination path.

Answer: The position vector of the point A that must lie at a fraction f on the destination path is A=[(5+11f)i^+(3+9f)j^]m .

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

Formula to calculate the position vector of the point A that must lie at a fraction f on the destination path is,

A=xi^+yj^

Substitute (5+11f)m for x and (3+9f)m for y to find A .

A=[(5+11f)i^+(3+9f)j^]m

Conclusion:

Therefore, position vector of point A in terms of fraction f is A=[(5+11f)i^+(3+9f)j^]m .

(b)

To determine

The position vector of point A for f=0 .

(b)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A for f=0 is A=[5i^+3j^]m .

Explanation of Solution

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m .

Substitute 0 for f in the above expression.

A=[(5+11×0)i^+(3+9×0)j^]m=[(5+0)i^+(3+0)j^]=5i^+3j^

Conclusion:

Therefore, position vector of point A for f=0 is A=[5i^+3j^]m .

(c)

To determine

Whether the position vector of point A for f=0 is reasonable.

(c)

Expert Solution
Check Mark

Explanation of Solution

Introduction: The position vector described the position of an object relative to a fixed reference point.

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

When the value of f is zero then the position vector of point A becomes,

A=[5i^+3j^]m

This position vector is similar to the initial position vector of the destination path. So, if the value of f is zero then the point A move to the initial point that really exist.

Conclusion:

Therefore, position vector of point A for f=0 is reasonable because the position of point A is shift to the starting point of the destination path.

(d)

To determine

The position vector of point A for f=1 .

(d)

Expert Solution
Check Mark

Answer to Problem 46P

The position vector of point A for f=1 is 16i^+12j^ .

Explanation of Solution

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

Substitute 1 for f in the above expression.

A=[(5+11×1)i^+(3+9×1)j^]m=[(5+11)i^+(3+9)j^]=16i^+12j^

Conclusion:

Therefore, position vector of point A for f=1 is 16i^+12j^ .

(e)

To determine

Whether the position vector of point A for f=1 is reasonable.

(e)

Expert Solution
Check Mark

Explanation of Solution

Introduction: The position vector described the position of an object relative to a fixed reference point.

Given information:

The initial position vector is (5i^+3j^)m and final position vector is (16i^+12j^)m .

The position vector of point A in terms of fraction f is,

A=[(5+11f)i^+(3+9f)j^]m

When the value of f is 1 , then the position vector of point A becomes,

A=16i^+12j^

This position vector is similar to the final position vector of the destination path. So, if the value of f is 1 then the point A move to the final point that really exist.

Conclusion:

Therefore, position vector of point A for f=1 is reasonable because the position of point A is shifted to the final point of the destination path.

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Chapter 3 Solutions

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

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