The position of a squirrel running in a park is given by r⃗=[(0.280m/s)t+(0.0360m/s2)t2]i^+(0.0190m/s3)t3j^.

At  t  = 4.00  s , what is the magnitude of the squirrel's velocity?
Express your answer with the appropriate units.

To find the magnitude of the squirrel's velocity at \( t = 4.00 \, \text{s} \), we need to first compute the velocity vector, which is the derivative of the position vector \( \mathbf{r}(t) \) with respect to time.

The position vector is given by:
\[
\mathbf{r}(t) = \left[ (0.280 \, \text{m/s})t + (0.0360 \, \text{m/s}^2)t^2 \right] \hat{i} + (0.0190 \, \text{m/s}^3)t^3 \hat{j}
\]

### Step 1: Find the velocity vector
The velocity vector \( \mathbf{v}(t) \) is the time derivative of the position vector \( \mathbf{r}(t) \):
\[
\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}
\]

#### Derivative of the \( \hat{i} \)-component:
\[
v_x(t) = \frac{d}{dt} \left[ (0.280 \, \text{m/s})t + (0.0360 \, \text{m/s}^2)t^2 \right] = 0.280 \, \text{m/s} + 2(0.0360 \, \text{m/s}^2)t
\]
\[
v_x(t) = 0.280 \, \text{m/s} + 0.0720 \, \text{m/s}^2 \cdot t
\]

#### Derivative of the \( \hat{j} \)-component:
\[
v_y(t) = \frac{d}{dt} \left[ (0.0190 \, \text{m/s}^3)t^3 \right] = 3(0.0190 \, \text{m/s}^3)t^2
\]
\[
v_y(t) = 0.0570 \, \text{m/s}^3 \cdot t^2
\]

### Step 2: Evaluate the velocity components at \( t = 4.00 \, \text{s} \)

#### \( v_x(4.00 \, \text{s}) \):
\[
v_x(4.00 \, \text{s}) = 0.280 \, \text{m/s} + 0.0720 \, \text{m/s}^2 \cdot 4.00 \, \text{s}
\]
\[
v_x(4.00 \, \text{s}) = 0.280 \, \text{m/s} + 0.288 \, \text{m/s} = 0.568 \, \text{m/s}
\]

#### \( v_y(4.00 \, \text{s}) \):
\[
v_y(4.00 \, \text{s}) = 0.0570 \, \text{m/s}^3 \cdot (4.00 \, \text{s})^2
\]
\[
v_y(4.00 \, \text{s}) = 0.0570 \, \text{m/s}^3 \cdot 16.00 = 0.912 \, \text{m/s}
\]

### Step 3: Find the magnitude of the velocity vector

The magnitude of the velocity vector is given by:
\[
|\mathbf{v}(t)| = \sqrt{v_x^2 + v_y^2}
\]
Substitute the values of \( v_x(4.00 \, \text{s}) \) and \( v_y(4.00 \, \text{s}) \):
\[
|\mathbf{v}(4.00 \, \text{s})| = \sqrt{(0.568 \, \text{m/s})^2 + (0.912 \, \text{m/s})^2}
\]
\[
|\mathbf{v}(4.00 \, \text{s})| = \sqrt{0.322624 + 0.831744}
\]
\[
|\mathbf{v}(4.00 \, \text{s})| = \sqrt{1.154368} \approx 1.074 \, \text{m/s}
\]

### Final Answer:
The magnitude of the squirrel's velocity at \( t = 4.00 \, \text{s} \) is approximately:
\[
1.07 \, \text{m/s}
\]

Do you want further clarification on any of these steps? 

Here are 5 related questions you can explore:
1. How do you calculate acceleration from a position function?
2. What is the difference between speed and velocity?
3. How do you compute the direction of the velocity vector?
4. What is the significance of the time derivative in motion analysis?
5. How does one integrate velocity to find position?

**Tip**: Always remember that the magnitude of velocity is the scalar quantity known as speed, and it is independent of direction!

The position of a squirrel running in a park is given by r⃗=[(0.280m/s)t+(0.0360m/s2)t2]i^+(0.0190m/s3)t3j^. At t = 4.00 s, what is the magnitude of the squirrel's velocity? Express your answer with the appropriate units.

Solve a kinematics problem involving a squirrel's motion in a park. This problem demonstrates how to find the velocity's magnitude by differentiating a position function and using vector analysis. Suitable for Grades 11-12 students, it covers calculus concepts and vector magnitude calculations.

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