Q3E - University Physics with Modern Physics

Question

A web page designer creates an animation in which a dot on a computer screen has position $\overrightarrow{\mathbf{r}}\mathbf{=}\mathbf{[}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{c}\mathbf{m}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{.}\mathbf{5}\mathbf{ }\mathbf{c}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}\mathbf{)}{\mathbf{t}}^{\mathbf{2}}\mathbf{]}\hat{\mathbf{i}}\mathbf{+}\mathbf{(}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{c}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathit{t}\hat{\mathbf{j}}$. 

(a) Find the magnitude and direction of the dot’s average velocity between $\mathit{t}\mathbf{=}\mathbf{0}$ and $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$.

(b) Find the magnitude and direction of the instantaneous velocity at $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{s}$, $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{s}$, and $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$. 

(c) Sketch the dot’s trajectory from $\mathit{t}\mathbf{=}\mathbf{0}\mathbf{ }\mathit{s}$ to $\mathit{t}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{ }\mathit{s}$, and show the velocities calculated in part (b).

Step-by-Step Solution

Answer

1. The magnitude and direction of average velocity are $7.1 \mathrm{cm}/\mathrm{s}$ and $45°$ respectively.
2. the instantaneous velocities for time intervals 0 to 2 seconds are $5 \mathrm{cm}/\mathrm{s}, 7.1 \mathrm{cm}/\mathrm{s}, 11.2 \mathrm{cm}/\mathrm{s}$ respectively and the direction of the velocities is $90°, 45°,$ and $26.6°$ respectively.
3. The sketch for the trajectory of the dot is shown as,

Step 1: Identification of given data

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, $t_2=2 s$.
• The position of the dot is $\overrightarrow{r}=[4.0 cm+(2.5 cm/s^2)t^2]\hat{i}+(5.0 cm/s)t\hat{j}$

Step 2: Concept/Significance of average velocity.

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.

Step 3: (a) Determination of the magnitude and direction of the dot’s average velocity between and

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

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

 

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

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

 

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

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

 

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.

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

 

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,

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

 

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,

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

 

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

Step 4: (b) Determination of the magnitude and direction of the instantaneous velocity at t = 0   s and t = 2 . 0   s

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

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

 

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

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

 

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

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

 

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

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

 

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,

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

 

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,

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

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

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

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

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

 

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

Step 4: (b) sketch of the dot’s trajectory from t = 0   s to t = 2 . 0   s , the velocities calculated in part b

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