Chapter 3, Problem 33P

In Figure P3.33, the line segment represents a path from the point with position vector $(5\hat{i}+3\hat{j})\text{m}$ to the point with location $(16\hat{i}+12\hat{j})\text{m}$. Point Ⓐ is along this path, a fraction f of the way to the destination. (a) Find the position vector of point Ⓐ in terms of f. (b) Evaluate the expression from part (a) for f = 0. (c) Explain whether the result in part (b) is reasonable. (d) Evaluate the expression for f = 1. (c) Explain whether the result in part (d) is reasonable.

Figure P3.33 Point Ⓐ is a fraction f of the distance from the initial point (5,3) to the final point (16, 12).

To determine

The position vector of point $\text{A}$ in terms of fraction $f$.

Answer to Problem 33P

The position vector of point $\text{A}$ in terms of fraction $f$ is $\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$.

Explanation of Solution

Section 1:

To determine: The horizontal position of point $\text{A}$ 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 $\left(5+11f\right)\text{m}$.

Explanation:

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

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

$x=x_1+f\left(x_2-x_1\right)$

• $x$ is the position of the point $\text{A}$ in x-direction.
• $x_1$ is the initial position in x-direction.
• $x_2$ is the final position in x-direction.
• $f$ is the fraction of the destination path.

Substitute $5\text{m}$ for $x_1$ and $16\text{m}$ for $x_2$.

$\begin{array}{c}x=5\text{m}+f\left(16\text{m}-5\text{m}\right)\\ =5\text{m}+f\left(11\text{m}\right)\\ =\left(5+11f\right)\text{m}\end{array}$

Section 2:

To determine: The vertical position of point $\text{A}$ that must be at a fraction $f$ of the destination.

Answer: The vertical position of point $\text{A}$ that must be at a fraction $f$ of the destination is $\left(3+9f\right)\text{m}$.

Explanation:

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

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

$y=y_1+f\left(y_2-y_1\right)$

• $y$ is the position of the point $A$ in y-direction.
• $y_1$ is the initial position in y-direction.
• $y_2$ is the final position in y-direction.
• $f$ is the fraction of the destination path.

Substitute $3\text{m}$ for $y_1$ and $12\text{m}$ for $y_2$.

$\begin{array}{c}y=3\text{m}+f\left(13\text{m}-3\text{m}\right)\\ =3\text{m}+f\left(9m\right)\\ =\left(3+9f\right)\text{m}\end{array}$

Section 3:

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

Answer: The position vector of the point $\text{A}$ that must lie at a fraction $f$ on the destination path is $\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$.

Explanation:

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

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

$\overrightarrow{\text{A}}=x\widehat{i}+y\widehat{j}$

Substitute $\left(5+11f\right)\text{m}$ for $x$ and $\left(3+9f\right)\text{m}$ for $y$ to find $\overrightarrow{\text{A}}$.

$\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$

Conclusion:

Therefore, position vector of point $\text{A}$ in terms of fraction $f$ is $\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$.

To determine

The position vector of point $\text{A}$ for $f=0$.

Answer to Problem 33P

The position vector of point $\text{A}$ for $f=0$ is $\overrightarrow{\text{A}}=\left[5\widehat{i}+3\widehat{j}\right]\text{m}$.

Explanation of Solution

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

The position vector of point $\text{A}$ in terms of fraction $f$ is,

$\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$.

Substitute $0$ for $f$ in the above expression.

$\begin{array}{c}\overrightarrow{\text{A}}=\left[\left(5+11\times 0\right)\widehat{i}+\left(3+9\times 0\right)\widehat{j}\right]\text{m}\\ \text{=}\left[\left(5+0\right)\widehat{i}+\left(3+0\right)\widehat{j}\right]\\ =5\widehat{i}+3\widehat{j}\end{array}$

Conclusion:

Therefore, position vector of point $A$ for $f=0$ is $\overrightarrow{\text{A}}=\left[5\widehat{i}+3\widehat{j}\right]\text{m}$.

To determine

Whether the position vector of point $\text{A}$ for $f=0$ is reasonable.

Explanation of Solution

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

Explanation:

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

The position vector of point $\text{A}$ in terms of fraction $f$ is,

$\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$

When the value of $f$ is zero then the position vector of point $\text{A}$ becomes,

$\overrightarrow{\text{A}}=\left[5\widehat{i}+3\widehat{j}\right]\text{m}$

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 $\text{A}$ move to the initial point that really exist.

Conclusion:

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

To determine

The position vector of point $\text{A}$ for $f=1$.

Answer to Problem 33P

The position vector of point $\text{A}$ for $f=1$ is $16\widehat{i}+12\widehat{j}$.

Explanation of Solution

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

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

$\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$

Substitute $1$ for $f$ in the above expression.

$\begin{array}{c}\overrightarrow{\text{A}}=\left[\left(5+11\times 1\right)\widehat{i}+\left(3+9\times 1\right)\widehat{j}\right]\text{m}\\ \text{=}\left[\left(5+11\right)\widehat{i}+\left(3+9\right)\widehat{j}\right]\\ =16\widehat{i}+12\widehat{j}\end{array}$

Conclusion:

Therefore, position vector of point $\text{A}$ for $f=1$ is $16\widehat{i}+12\widehat{j}$.

To determine

Whether the position vector of point $\text{A}$ for $f=1$ is reasonable.

Explanation of Solution

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

Explanation:

Given information:

The initial position vector is $\left(5\widehat{i}+3\widehat{j}\right)\text{m}$ and final position vector is $\left(16\widehat{i}+12\widehat{j}\right)\text{m}$.

The position vector of point $\text{A}$ in terms of fraction $f$ is,

$\overrightarrow{\text{A}}=\left[\left(5+11f\right)\widehat{i}+\left(3+9f\right)\widehat{j}\right]\text{m}$

When the value of $f$ is $1$, then the position vector of point $\text{A}$ becomes,

$\overrightarrow{A}=16\widehat{i}+12\widehat{j}$

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 $\text{A}$ move to the final point that really exist.

Conclusion:

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