Chapter 2, Problem 4P

(a)

To determine

The displacement of the car between $3.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$.

Answer to Problem 4P

The displacement of the car between $3.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$ is $8\text{km}$ north of its position at $3\text{P}\text{.M}$.

Explanation of Solution

Take south direction as $+x$ direction.

The vector diagram of the displacement of the car at different time is shown below.

From the figure at $3.00\text{P}\text{.M}$ position of the object is $20\text{km}$ from starting point and at $6.00\text{P}\text{.M}$ position of object is $12\text{km}$.

Write the expression for the displacement of the car between $3.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$.

$\Delta x=x_3-x_1$ (I)

Here, $\Delta x$ is the displacement of the car , $x_1$ is the position of the car at $3.00\text{P}\text{.M}$ and $x_3$ is the position of the car at $6.00\text{P}\text{.M}$.

Vector diagram of $-x_{1,}{x}_3{\text{and x}}_3-x_1$ is shown below.

Conclusion:

Substitute $12\text{km}$ for $x_3$ and $20\text{km}$ for $x_1$ in equation (I) to get $\Delta x$.

$\begin{array}{c}\Delta x=12\text{km}-20\text{km}\\ \text{=}-8\text{km}\end{array}$

Since $+x$ is taken as south direction, negative sign for $\Delta x$ indicate that displacement of the car is towards north direction.

Therefore, the displacement of the car between $3.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$ is $8\text{km}$ north of its position at $3\text{P}\text{.M}$.

(b)

To determine

The displacement of the car from starting point to location at $4.00\text{P}\text{.M}$.

Answer to Problem 4P

The displacement of the car from starting point to location at $4.00\text{P}\text{.M}$ is $116\text{km}$ south of the starting point.

Explanation of Solution

Take south direction as $+x$ direction.

The vector diagram of the displacement of the car at different time is shown below.

From the figure at $4.00\text{P}\text{.M}$, the position of the car is $96\text{km}$ south from position of the car at $3.00\text{P}\text{.M}$.

Write the expression for the displacement of the car from starting point to position of the car at $4.00\text{P}\text{.M}$.

$\Delta x=x_1+x_2$ (II)

Here, $\Delta x$ is the displacement of the car , $x_1$ is the position of the car at $3.00\text{P}\text{.M}$ and $x_2$ is the position of the car at $4.00\text{P}\text{.M}$.

Vector diagram of $x_1$ , $x_2$ and $x_1+x_2$ is shown below.

Conclusion:

Substitute $20\text{km}$ for $x_1$ and $96\text{km}$ for $x_2$ in equation (II) to get $\Delta x$.

$\begin{array}{c}\Delta x=20\text{km}+96\text{km}\\ \text{=}116\text{km}\end{array}$

Therefore, the displacement of the car from starting point to location at $4.00\text{P}\text{.M}$ is $116\text{km}$ south of the starting point.

(c)

To determine

The displacement of the car between $4.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$.

Answer to Problem 4P

The displacement of the car between $4.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$ is $104\text{km}$ north of its position at $4\text{P}\text{.M}$.

Explanation of Solution

Take south direction as $+x$ direction.

The vector diagram of the displacement of the car at different time is shown below.

From the figure at $4.00\text{P}\text{.M}$, position of the object is $96\text{km}$ south from position of the car at $3.00\text{P}\text{.M}$ and at $6.00\text{P}\text{.M}$ position of object is $12\text{km}$ south from starting point.

Write the expression for the displacement of the car between $4.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$.

$\Delta x=x_3-\left(x_1+x_2\right)$ (III)

Here, $\Delta x$ is the displacement of the car , $x_1$ is the position of the car at $3.00\text{P}\text{.M}$ and $x_3$ is the position of the car at $6.00\text{P}\text{.M}$.

Vector diagram of $-\left(x_1+x_2\right),x_3{\text{and x}}_3-\left(x_1+x_2\right)$ is shown below.

Conclusion:

Substitute $12\text{km}$ for $x_3$, $96\text{km}$ for $x_2$ and $20\text{km}$ for $x_1$ in equation (III) to get $\Delta x$.

$\begin{array}{c}\Delta x=12\text{km}-\left(20\text{km}+96\text{km}\right)\\ =-104\text{km}\end{array}$

Since $+x$ is taken as south direction, negative sign for $\Delta x$ indicate that displacement of the car is towards north direction.

Therefore, the displacement of the car between $4.00\text{P}\text{.M}$ and $6.00\text{P}\text{.M}$ is $104\text{km}$ north of its position at $4\text{P}\text{.M}$.