A web page designer creates a animation in which a dot in the computer screen has position r=2.0 cm +((1.5 m/s^2 )t^2)i+(3.0 cm/s)tj. a) Find the magnitude and directions of the dots average velocity between t=0 and t=3s. b) Find the magnitude and direction of the instantaneous velocity at t=0, t=1s, and t=2s

Given the position vector of the dot as a function of time:

\[
\mathbf{r}(t) = 2.0 \, \text{cm} + \left(1.5 \, \frac{\text{m}}{\text{s}^2}\right) t^2 \, \mathbf{i} + \left(3.0 \, \frac{\text{cm}}{\text{s}}\right) t \, \mathbf{j}
\]

First, let's break down the vector into its components:

\[
\mathbf{r}(t) = (2.0 \, \text{cm} + 1.5 \, \frac{\text{m}}{\text{s}^2} \cdot t^2) \, \mathbf{i} + (3.0 \, \frac{\text{cm}}{\text{s}} \cdot t) \, \mathbf{j}
\]

Since 1.5 m/s² = 150 cm/s² (to keep units consistent), the position vector can be rewritten as:

\[
\mathbf{r}(t) = (2.0 \, \text{cm} + 150 \, \frac{\text{cm}}{\text{s}^2} \cdot t^2) \, \mathbf{i} + (3.0 \, \frac{\text{cm}}{\text{s}} \cdot t) \, \mathbf{j}
\]

### Part (a): Magnitude and Direction of Average Velocity Between \( t = 0 \) and \( t = 3 \, \text{s} \)

The average velocity \(\mathbf{v}_{\text{avg}}\) is given by:

\[
\mathbf{v}_{\text{avg}} = \frac{\mathbf{r}(t_2) - \mathbf{r}(t_1)}{t_2 - t_1}
\]

Here, \( t_1 = 0 \, \text{s} \) and \( t_2 = 3 \, \text{s} \).

First, calculate \( \mathbf{r}(0) \):

\[
\mathbf{r}(0) = 2.0 \, \text{cm} \, \mathbf{i} + 0 \, \mathbf{j} = 2.0 \, \mathbf{i} \, \text{cm}
\]

Next, calculate \( \mathbf{r}(3) \):

\[
\mathbf{r}(3) = (2.0 + 150 \cdot 3^2) \, \mathbf{i} + (3.0 \cdot 3) \, \mathbf{j}
\]
\[
\mathbf{r}(3) = (2.0 + 1350) \, \mathbf{i} + 9.0 \, \mathbf{j} = 1352.0 \, \mathbf{i} \, \text{cm} + 9.0 \, \mathbf{j} \, \text{cm}
\]

Now, find \( \mathbf{v}_{\text{avg}} \):

\[
\mathbf{v}_{\text{avg}} = \frac{1352.0 \, \mathbf{i} + 9.0 \, \mathbf{j} - 2.0 \, \mathbf{i} - 0 \, \mathbf{j}}{3 \, \text{s}} = \frac{1350.0 \, \mathbf{i} + 9.0 \, \mathbf{j}}{3 \, \text{s}}
\]

\[
\mathbf{v}_{\text{avg}} = 450.0 \, \frac{\text{cm}}{\text{s}} \, \mathbf{i} + 3.0 \, \frac{\text{cm}}{\text{s}} \, \mathbf{j}
\]

To find the magnitude of the average velocity:

\[
|\mathbf{v}_{\text{avg}}| = \sqrt{(450.0)^2 + (3.0)^2} \, \frac{\text{cm}}{\text{s}} 
\]

\[
|\mathbf{v}_{\text{avg}}| \approx 450.01 \, \frac{\text{cm}}{\text{s}} 
\]

The direction (angle \(\theta\)) with respect to the positive x-axis is given by:

\[
\theta = \tan^{-1}\left(\frac{3.0}{450.0}\right)
\]

\[
\theta \approx 0.38^\circ 
\]

### Part (b): Instantaneous Velocity at \( t = 0 \, \text{s} \), \( t = 1 \, \text{s} \), and \( t = 2 \, \text{s} \)

The instantaneous velocity \(\mathbf{v}(t)\) is the derivative of the position vector \(\mathbf{r}(t)\) with respect to time \(t\):

\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt} = \frac{d}{dt} \left( (2.0 + 150 \, t^2) \, \mathbf{i} + (3.0 \, t) \, \mathbf{j} \right)
\]

\[
\mathbf{v}(t) = 300 \, t \, \mathbf{i} + 3.0 \, \mathbf{j} \, \frac{\text{cm}}{\text{s}}
\]

Now, calculate the instantaneous velocity at the specified times:

1. **At \( t = 0 \, \text{s} \):**
   \[
   \mathbf{v}(0) = 300 \cdot 0 \, \mathbf{i} + 3.0 \, \mathbf{j} = 3.0 \, \mathbf{j} \, \frac{\text{cm}}{\text{s}}
   \]
   - Magnitude: \( |\mathbf{v}(0)| = 3.0 \, \frac{\text{cm}}{\text{s}} \)
   - Direction: Along the positive y-axis (90°).

2. **At \( t = 1 \, \text{s} \):**
   \[
   \mathbf{v}(1) = 300 \cdot 1 \, \mathbf{i} + 3.0 \, \mathbf{j} = 300 \, \mathbf{i} + 3.0 \, \mathbf{j} \, \frac{\text{cm}}{\text{s}}
   \]
   - Magnitude: 
   \[
   |\mathbf{v}(1)| = \sqrt{(300)^2 + (3.0)^2} \approx 300.01 \, \frac{\text{cm}}{\text{s}}
   \]
   - Direction:
   \[
   \theta = \tan^{-1}\left(\frac{3.0}{300}\right) \approx 0.57^\circ
   \]

3. **At \( t = 2 \, \text{s} \):**
   \[
   \mathbf{v}(2) = 300 \cdot 2 \, \mathbf{i} + 3.0 \, \mathbf{j} = 600 \, \mathbf{i} + 3.0 \, \mathbf{j} \, \frac{\text{cm}}{\text{s}}
   \]
   - Magnitude: 
   \[
   |\mathbf{v}(2)| = \sqrt{(600)^2 + (3.0)^2} \approx 600.01 \, \frac{\text{cm}}{\text{s}}
   \]
   - Direction:
   \[
   \theta = \tan^{-1}\left(\frac{3.0}{600}\right) \approx 0.29^\circ
   \]

### Summary:
- **Average velocity:** \(450.01 \, \text{cm/s}\) at \(0.38^\circ\) to the x-axis.
- **Instantaneous velocities:**
  - At \( t = 0 \, \text{s} \): \(3.0 \, \text{cm/s}\) at \(90^\circ\) to the x-axis.
  - At \( t = 1 \, \text{s} \): \(300.01 \, \text{cm/s}\) at \(0.57^\circ\) to the x-axis.
  - At \( t = 2 \, \text{s} \): \(600.01 \, \text{cm/s}\) at \(0.29^\circ\) to the x-axis.

Would you like to see more details or have any questions?

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### Follow-up Questions:
1. How would the average velocity change if the time interval were different?
2. What is the significance of the instantaneous velocity direction at each point?
3. How does the acceleration affect the velocity components over time?
4. Can the dot ever move exactly along the x-axis? Under what conditions?
5. How would the results change if the \(j\) component of the position function had a different coefficient?

### Tip:
Remember that the magnitude of velocity is the rate at which the position changes with time, while the direction indicates the orientation of the motion.

A web page designer creates an animation in which a dot on the computer screen has a position r=2.0 cm +((1.5 m/s^2)t^2)i+(3.0 cm/s)tj. a) Find the magnitude and directions of the dot's average velocity between t=0 and t=3s. b) Find the magnitude and direction of the instantaneous velocity at t=0, t=1s, and t=2s.

Learn how to calculate the average and instantaneous velocities of a dot in a kinematic animation using vector calculus. The problem involves finding magnitudes and directions at specific time intervals, suitable for students in Grades 11-12.

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