91Ó°ÊÓ

The given data can be listed below,

• The initial time of a web designer is,$t_1=0$
• The final time of the web designer is, $t_2=2s$.
• The position of the dot is$\stackrel{→}{r}=[4.0cm+(2.5cm/s^2)t^2]\hat{i}+(5.0cm/s)t\hat{j}$

The average velocity of a body traveling in a certain direction is defined as the ratio of the body's displacement to the entire journey duration.

The position of the dot at the time$t=0s$ is given by,

$$\begin{gathered} r_1=[\left(4.0cm\right)+(2.5cm/s^2){\left(0s\right)}^2]\hat{i}+\left((5.0cm/s)\left(0s\right)\right)\hat{j} \\ =\left(4.0cm\right)\hat{i} \end{gathered}$$

The position of the dot at the time$t=1s$ is given by,

role="math" localid="1664879760372" $$\begin{gathered} r_2=[\left(4.0cm\right)+(2.5cm/s^2){\left(1s\right)}^2]\hat{i}+\left((5.0cm/s)\left(1s\right)\right)\hat{j} \\ =\left(6.5cm\right)\hat{i}+\left(5.0cm\right)\hat{j} \end{gathered}$$

The position of the dot at the time is given by,

role="math" localid="1664879772087" $$\begin{gathered} r_3=[\left(4.0cm\right)+(2.5cm/s^2){\left(2s\right)}^2]\hat{i}+\left((5.0cm/s)\left(2s\right)\right)\hat{j} \\ =\left(14cm\right)\hat{i}+\left(10cm\right)\hat{j} \end{gathered}$$

The average velocity of the dot is given by,

$v_{avg}=\frac{r_3-r_1}{t_3-t_1}$

Here,$r_3$is the final position of the dot at time 2 s,$r$is the initial position of the dot at time 0 s,$t_3$is the final time and$t_1$is the initial time of the dot.

Substitute all the values in the above equation.

$$\begin{gathered} v_{avg}=\frac{\left(14cm\right)\hat{i}+\left(10cm\right)\hat{j}-\left(4cm\right)\hat{i}}{\left(2-0\right)s} \\ =\left(5cm/s\right)\hat{i}+\left(5cm/s\right)\hat{j} \end{gathered}$$

The magnitude of the average velocity is calculated as,

$$\begin{gathered} \left|v_{avg}\right|=\sqrt{{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2+{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2} \\ =7.1\mathrm{cm}/\mathrm{s} \end{gathered}$$

The direction of average velocity is given by,

$$\begin{gathered} \tan \mathrm{Î\pm }=\frac{v_{avg,j}}{v_{avg,i}} \\ \mathrm{Î\pm }={\tan }^{-1}\left(\frac{v_{avg,j}}{v_{avg,i}}\right) \end{gathered}$$

Here,$v_{avg,j}$is the y-component of average velocity and$v_{avg,i}$is the x-component of average velocity.

Substitute all the values in the above,

$$\begin{gathered} \mathrm{Î\pm }={\tan }^{-1}\left(\frac{5\mathrm{cm}/\mathrm{s}}{5\mathrm{cm}/\mathrm{s}}\right) \\ =45° \end{gathered}$$

Thus, the magnitude and direction of average velocity are $7.1\mathrm{cm}/\mathrm{s}$and $45°$respectively.

The instantaneous velocity of the dot is given by differentiating position with respect to time.

$v=\frac{dr}{dt}$

Substitute the values in the above equation

$$\begin{gathered} v=\frac{d}{dt}\left(\left[4.0cm+\left(2.5cm/s^2\right)t^2\right]\hat{i}+\left(5.0cm/s\right)t\hat{j}\right) \\ =\left(2\left(2.5cm/s^2\right)t\right)\hat{i}+\left(5cm/s\right)\hat{j} \\ =\left(5tcm/s\right)\hat{i}+5\hat{j}cm/s \end{gathered}$$

The velocity of the dot at the time$t=0s$ is given by,

$$\begin{gathered} v_1=\left(5cm/s^{2}t\right)\hat{i}+\hat{j}5cm/s \\ =\left(5cm/s\left(0\right)\right)\hat{i}+\hat{j}5cm/s \\ =\left(5cm/s\right)\hat{j} \end{gathered}$$

The velocity of the dot at the time$t=0s$ is given by,

$$\begin{gathered} v_2=\left(\left(5.0cm/s^2\right)\left(1s\right)\right)\hat{i}+\left(5.0cm/s\right)\hat{j} \\ =\left(5.0cm/s\right)\hat{i}+\left(5.0cm/s\right)\hat{j} \end{gathered}$$

The velocity of the dot at the time$t=2s$ is given by,

$$\begin{gathered} v_3=\left(\left(5.0cm/s^2\right)\left(2s\right)\right)\hat{i}+\left(5.0cm/s\right)\hat{j} \\ =\left(10cm/s\right)\hat{i}+\left(5.0cm/s\right)\hat{j} \end{gathered}$$

The magnitude of the instantaneous velocity$v_1$ is calculated as,

$$\begin{gathered} \left|v_1\right|=\sqrt{{\left(0\right)}^2+{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2} \\ =5\mathrm{cm}/\mathrm{s} \end{gathered}$$

The direction of the dot for$v_1$ is given by,

$$\begin{gathered} \mathrm{Î\pm }={\tan }^{-1}\left(\frac{5cm/s}{0}\right) \\ =90° \end{gathered}$$

The magnitude of the instantaneous velocity$v_2$ is calculated as,

$$\begin{gathered} \left|v_2\right|=\sqrt{{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2+{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2} \\ =7.1\mathrm{cm}/\mathrm{s} \end{gathered}$$

The direction of the dot for $v_2$is given by,

$$\begin{gathered} \mathrm{Î\pm }={\tan }^{-1}\left(\frac{5\mathrm{cm}/\mathrm{s}}{5\mathrm{cm}/\mathrm{s}}\right) \\ =45° \end{gathered}$$

The magnitude of the instantaneous velocity$v_3$is calculated as,

role="math" localid="1664881150047" $$\begin{gathered} \left|v_3\right|=\sqrt{{\left(10\mathrm{cm}/\mathrm{s}\right)}^2+{\left(5.0\mathrm{cm}/\mathrm{s}\right)}^2} \\ =11.2\mathrm{cm}/\mathrm{s} \end{gathered}$$

The direction of the dot for$v_3$is given by,

$$\begin{gathered} \mathrm{Î\pm }={\tan }^{-1}\left(\frac{5\mathrm{cm}/\mathrm{s}}{10\mathrm{cm}/\mathrm{s}}\right) \\ =26.6° \end{gathered}$$

Thus, the instantaneous velocities for time intervals 0 to 2 seconds is $5cm/s,7.1cm/s,11.2cm/s$respectively and the direction of the velocities is $90°,45°,$and $26.6°$respectively.

The trajectory of the dot is shown in the graph below where the slopes of the graph give the instantaneous velocities.