Chapter 3, Problem 2P

(a)

To determine

The average velocity during time interval $t=2.00\text{ s}$ to $t=4.00\text{ s}$.

Answer to Problem 2P

The average velocity during time interval $t=2.00\text{ s}$ to $t=4.00\text{ s}$ is $\left(\text{1}\text{.00 m}\right)\widehat{i}+\left(\text{0}\text{.75 m}\right)\widehat{j}$.

Explanation of Solution

Section 1:

To determine: The position vector $\overrightarrow{r_{\text{i}}}$.

Answer: The position vector $\overrightarrow{r_{\text{i}}}$ is $\left(\text{3}\text{.00 m}\right)\widehat{i}+\left(\text{1}\text{.5 m}\right)\widehat{j}$.

Given information:

The value of $a$ is $\text{1 }\text{m}/\text{s}$, the value of $b$ is $\text{1}\text{.00 m}$, the value of $c$ is $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ and the value of $d$ is $\text{1}\text{.00 m}$.

The position vector at time $t=2.0\text{s}$ is written as,

$\begin{array}{c}\overrightarrow{r_{\text{i}}}=x\left(t\right)\widehat{i}+y\left(t\right)\widehat{j}\\ =\left(at+b\right)\widehat{i}+\left(c{t}^2+d\right)\widehat{j}\end{array}$

Substitute $2.00\text{ s}$ for $t$, $\text{1 }\text{m}/\text{s}$ for $a$, $\text{1}\text{.00 m}$ for $b$, $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ for $c$ and $\text{1}\text{.00 m}$ for $d$ in above equation to find $\overrightarrow{r_{\text{i}}}$.

$\begin{array}{c}\overrightarrow{r_{\text{i}}}=\left(\left(\text{1 }\text{m}/\text{s}\right)\left(2.00\text{ s}\right)+\left(\text{1}\text{.00 m}\right)\right)\widehat{i}+\left(\left(\text{0}\text{.125 }\text{m}/{\text{s}}^2\right){\left(2.00\text{ s}\right)}^2+\left(\text{1}\text{.00 m}\right)\right)\widehat{j}\\ =\left(\text{3}\text{.00 m}\right)\widehat{i}+\left(\text{1}\text{.5 m}\right)\widehat{j}\end{array}$

Section 2:

To determine: The position vector $\overrightarrow{r_{\text{f}}}$.

Answer: The position vector $\overrightarrow{r_{\text{f}}}$ is $\left(\text{5}\text{.00 m}\right)\widehat{i}+\left(\text{3}\text{.00 m}\right)\widehat{j}$.

Given information:

The value of $a$ is $\text{1 }\text{m}/\text{s}$, the value of $b$ is $\text{1}\text{.00 m}$, the value of $c$ is $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ and the value of $d$ is $\text{1}\text{.00 m}$.

The position vector at time $t=4.0\text{s}$ is written as,

$\overrightarrow{r_{\text{f}}}=\left(at+b\right)\widehat{i}+\left(c{t}^2+d\right)\widehat{j}$

Substitute $4.00\text{ s}$ for $t$, $\text{1 }\text{m}/\text{s}$ for $a$, $\text{1}\text{.00 m}$ for $b$, $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ for $c$ and $\text{1}\text{.00 m}$ for $d$ in above equation to find $\overrightarrow{r_{\text{i}}}$.

$\begin{array}{c}\overrightarrow{r_{\text{f}}}=\left(\left(\text{1 }\text{m}/\text{s}\right)\left(4.00\text{ s}\right)+\left(\text{1}\text{.00 m}\right)\right)\widehat{i}+\left(\left(\text{0}\text{.125 }\text{m}/{\text{s}}^2\right){\left(4.00\text{ s}\right)}^2+\left(\text{1}\text{.00 m}\right)\right)\widehat{j}\\ =\left(\text{5}\text{.00 m}\right)\widehat{i}+\left(\text{3}\text{.00 m}\right)\widehat{j}\end{array}$

Section 3:

To determine: The average velocity during time interval $t=2.00\text{ s}$ to $t=4.00\text{ s}$.

Answer: The average velocity during time interval $t=2.00\text{ s}$ to $t=4.00\text{ s}$ is $\left(\text{1}\text{.00 m}\right)\widehat{i}+\left(\text{0}\text{.75 m}\right)\widehat{j}$.

Given information:

The value of $a$ is $\text{1 }\text{m}/\text{s}$, the value of $b$ is $\text{1}\text{.00 m}$, the value of $c$ is $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ and the value of $d$ is $\text{1}\text{.00 m}$.

The formula to calculate average velocity is,

${\overrightarrow{v}}_{\text{avg}}=\frac{\overrightarrow{r_{\text{f}}}-\overrightarrow{r_{\text{i}}}}{t_{\text{f}}-t_{\text{i}}}$

Substitute $\left(\text{3}\text{.00 m}\right)\widehat{i}+\left(\text{1}\text{.5 m}\right)\widehat{j}$ for $\overrightarrow{r_{\text{i}}}$, $\left(\text{5}\text{.00 m}\right)\widehat{i}+\left(\text{3}\text{.00 m}\right)\widehat{j}$ for $\overrightarrow{r_{\text{f}}}$, $2.00\text{ s}$ for $t_{\text{i}}$ and $4.00\text{ s}$ for $t_{\text{f}}$ in above equation to find ${\overrightarrow{v}}_{\text{avg}}$.

$\begin{array}{c}{\overrightarrow{v}}_{\text{avg}}=\frac{\left(\left(\text{5}\text{.00 m}\right)\widehat{i}+\left(\text{3}\text{.00 m}\right)\widehat{j}\right)-\left(\left(\text{3}\text{.00 m}\right)\widehat{i}+\left(\text{1}\text{.5 m}\right)\widehat{j}\right)}{\left(4.00\text{ s}-2.00\text{ s}\right)}\\ =\frac{\left(5.00\text{m}-3.00\text{m}\right)\widehat{i}+\left(3.00\text{m}-1.5\text{m}\right)\widehat{j}}{}\\ =\frac{\left(\text{2}\text{.00 m}\right)\widehat{i}+\left(\text{1}\text{.5 m}\right)\widehat{j}}{\left(2.00\text{ s}\right)}\\ =\left(\text{1}\text{.00 m}\right)\widehat{i}+\left(\text{0}\text{.75 m}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the average velocity during time interval $t=2.00\text{ s}$ to $t=4.00\text{ s}$ is $\left(\text{1}\text{.00 m}\right)\widehat{i}+\left(\text{0}\text{.75 m}\right)\widehat{j}$.

(b)

To determine

The velocity and speed at time $t=2.00\text{ s}$.

Answer to Problem 2P

The velocity at time $t=2.00\text{ s}$ is $\left(\text{1 }\text{m}/\text{s}\right)\widehat{i}+\left(\text{0}\text{.5 }\text{m}/\text{s}\right)\widehat{j}$ and the speed at time $t=2.00\text{ s}$ is $1.11\text{m}/\text{s}$.

Explanation of Solution

Section 1:

To determine: The velocity at time $t=2.00\text{ s}$.

Answer: The velocity at time $t=2.00\text{ s}$ is $\left(\text{1 }\text{m}/\text{s}\right)\widehat{i}+\left(\text{0}\text{.5 }\text{m}/\text{s}\right)\widehat{j}$.

Given information:

The value of $a$ is $\text{1 }\text{m}/\text{s}$, the value of $b$ is $\text{1}\text{.00 m}$, the value of $c$ is $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ and the value of $d$ is $\text{1}\text{.00 m}$.

The formula to calculate velocity at time $t=2.00\text{ s}$ is,

$\begin{array}{c}\overrightarrow{v}=\underset{\Delta t\to 0}{\lim }\frac{\Delta \overrightarrow{r}}{\Delta \overrightarrow{t}}\\ =\frac{d\overrightarrow{r}}{dt}\\ =\frac{d\left(\left(\left(at+b\right)\widehat{i}+\left(c{t}^2+d\right)\widehat{j}\right)\right)}{dt}\\ =a\widehat{i}+\left(2ct\right)\widehat{j}\end{array}$

Substitute $2.00\text{ s}$ for $t$, $\text{1 }\text{m}/\text{s}$ for $a$ and $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ for $c$ in above equation to find $\overrightarrow{v}$.

$\begin{array}{c}\overrightarrow{v}=\left(\text{1 }\text{m}/\text{s}\right)\widehat{i}+\left(2\left(\text{0}\text{.125 }\text{m}/{\text{s}}^2\right)\left(2.00\text{ s}\right)\right)\widehat{j}\\ =\left(\text{1 }\text{m}/\text{s}\right)\widehat{i}+\left(\text{0}\text{.5 }\text{m}/\text{s}\right)\widehat{j}\end{array}$

Conclusion:

Therefore, the velocity at time $t=2.00\text{ s}$ is $\left(\text{1 }\text{m}/\text{s}\right)\widehat{i}+\left(\text{0}\text{.5 }\text{m}/\text{s}\right)\widehat{j}$.

Section 2:

To determine: The speed at time $t=2.00\text{ s}$.

Answer: The speed at time $t=2.00\text{ s}$ is $1.12\text{m}/\text{s}$.

Given information:

The value of $a$ is $\text{1 }\text{m}/\text{s}$, the value of $b$ is $\text{1}\text{.00 m}$, the value of $c$ is $\text{0}\text{.125 }\text{m}/{\text{s}}^2$ and the value of $d$ is $\text{1}\text{.00 m}$.

The formula to calculate speed at $t=2.00\text{ s}$ is,

$v=\left|\overrightarrow{v}\right|$

Calculate the magnitude of velocity as,

$\begin{array}{c}\left|\overrightarrow{v}\right|=\sqrt{{\left(\text{1 }\text{m}/\text{s}\right)}^2+{\left(\text{0}\text{.5 }\text{m}/\text{s}\right)}^2}\\ =1.12\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed during time interval $t=2.00\text{ s}$ is $1.12\text{m}/\text{s}$.