Chapter 3, Problem 4P

An ant crawls on the floor along the curved path shown in Figure P3.4. The ant’s positions and velocities are indicated for times t_i = 0 and t_f = 5.00 s. Determine the x- and y-components of the ant’s (a) displacement, (b) average velocity, and (c) average acceleration between the two times.

Figure P3.4

To determine

The $x$ and $y$ components of the ant’s displacement.

Answer to Problem 4P

The $x$ and $y$ components of the ant’s displacement are $0.300\text{m}$ and $0.200\text{m}$ respectively.

Explanation of Solution

Given info: The final $x$ coordinate of the ant’s position is $0.400\text{m}$, the initial $x$ coordinate of the ant’s position is $0.100\text{m}$, the final $y$ coordinate of the ant’s position is $0.350\text{m}$ and the initial $y$ coordinate of the ant’s position is $0.150\text{m}$.

Explanation:

The $x$ component of the ant’s displacement is,

$\Delta x=x_f-x_i$

• $x_f$ is the final $x$ coordinate
• $x_i$ is the initial $x$ coordinate

Substitute $0.100\text{m}$ for $x_i$ and $0.400\text{m}$ for $x_f$.

$\begin{array}{c}\Delta x=0.400\text{m}-0.100\text{m}\\ =\text{0}\text{.300}\text{m}\end{array}$

The $y$ component of the ant’s displacement is,

$\Delta y=y_f-y_i$

• $y_f$ is the final $y$ coordinate
• $y_i$ is the initial $y$ coordinate

Substitute $0.150\text{m}$ for $y_i$ and $0.350\text{m}$ for $y_f$.

$\begin{array}{c}\Delta y=0.350\text{m}-0.150\text{m}\\ =\text{0}\text{.200}\text{m}\end{array}$

Conclusion:

The $x$ and $y$ components of the ant’s displacement are $0.300\text{m}$ and $0.200\text{m}$ respectively.

To determine

The $x$ and $y$ components of the ant’s average velocity.

Answer to Problem 4P

The $x$ and $y$ components of the ant’s average velocity are $0.060\text{m/s}$ and $0.040\text{m/s}$ respectively.

Explanation of Solution

Given info: The $x$ component of the ant’s displacement is $0.300\text{m}$, the $y$ component of the ant’s displacement is $0.200\text{m}$ and the time taken is $5.00\text{s}$.

Explanation:

The $x$ component of the ant’s average velocity is,

$v_{av,x}=\frac{\Delta x}{\Delta t}$

• $\Delta x$ is the $x$ component of the displacement
• $\Delta t$ is the time taken

Substitute $0.300\text{m}$ for $\Delta x$ and $5.00\text{s}$ for $\Delta t$.

$\begin{array}{c}{v}_{av,x}=\frac{\text{0}\text{.300}\text{m}}{5.00\text{s}}\\ =0.060\text{m/s}\end{array}$

The $y$ component of the ant’s average velocity is,

$v_{av,y}=\frac{\Delta y}{\Delta t}$

• $\Delta y$ is the $y$ component of the displacement
• $\Delta t$ is the time taken

Substitute $0.200\text{m}$ for $\Delta y$ and $5.00\text{s}$ for $\Delta t$.

$\begin{array}{c}{v}_{av,y}=\frac{\text{0}\text{.200}\text{m}}{5.00\text{s}}\\ =0.040\text{m/s}\end{array}$

Conclusion:

The $x$ and $y$ components of the ant’s average velocity are $0.060\text{m/s}$ and $0.040\text{m/s}$ respectively.

To determine

The $x$ and $y$ components of the ant’s average acceleration.

Answer to Problem 4P

The $x$ and $y$ components of the ant’s average acceleration are $0.020{\text{m/s}}^2$ and $-0.020{\text{m/s}}^2$ respectively.

Explanation of Solution

Given info: The final $x$ coordinate of the ant’s velocity is $0.100\text{m/s}$, the initial $x$ coordinate of the ant’s velocity is $0\text{m/s}$, the final $y$ coordinate of the ant’s velocity is $0\text{m/s}$, the initial $y$ coordinate of the ant’s velocity is $0.100\text{m/s}$ and the time taken is $5.00\text{s}$.

Explanation:

The $x$ component of the ant’s average acceleration is,

$\begin{array}{c}{a}_{av,x}=\frac{\Delta v_x}{\Delta t}\\ =\frac{v_{x,f}-v_{x,i}}{\Delta t}\end{array}$

• $v_{x,f}$ is the $x$ component of the final velocity
• $v_{x,i}$ is the $x$ component of the initial velocity
• $\Delta t$ is the time taken

Substitute $0.100\text{m/s}$ for $v_{x,f}$, $0\text{m/s}$ for $v_{x,i}$ and $5.00\text{s}$ for $\Delta t$.

$\begin{array}{c}{a}_{av,x}=\frac{\text{0}\text{.100}\text{m/s}-0\text{m/s}}{5.00\text{s}}\\ =0.020{\text{m/s}}^2\end{array}$

The $y$ component of the ant’s average acceleration is,

$\begin{array}{c}{a}_{av,y}=\frac{\Delta v_y}{\Delta t}\\ =\frac{v_{y,f}-v_{y,i}}{\Delta t}\end{array}$

• $v_{y,f}$ is the $y$ component of the final velocity
• $v_{y,i}$ is the $y$ component of the initial velocity
• $\Delta t$ is the time taken

Substitute $0.100\text{m/s}$ for $v_{y,i}$, $0\text{m/s}$ for $v_{y,f}$ and $5.00\text{s}$ for $\Delta t$.

$\begin{array}{c}{a}_{av,y}=\frac{0\text{m/s}-\text{0}\text{.100}\text{m/s}}{5.00\text{s}}\\ =-0.020{\text{m/s}}^2\end{array}$

Conclusion:

The $x$ and $y$ components of the ant’s average acceleration are $0.020{\text{m/s}}^2$ and $-0.020{\text{m/s}}^2$ respectively.