The position of a particular particle as a function of time is given by r⃗ 
 = ( 9.60t⋅i^
 + 8.85j^
 - 1.00t2⋅k^
)m
, where t
 is in seconds.What is the average velocity of the particle between  t
 =1.00  s
  and  t
 =3.00  s
 ?
Express your answer in terms of the unit vectors  i^
 , j^
 , and  k^
 .

To find the average velocity of the particle between \( t = 1.00 \, \text{s} \) and \( t = 3.00 \, \text{s} \), we first need to determine the position vector \( \vec{r}(t) \) at these two times and then use the formula for average velocity.

### Given Position Vector
The position of the particle as a function of time is given by:
\[
\vec{r}(t) = (9.60t \, \hat{i} + 8.85 \, \hat{j} - 1.00t^2 \, \hat{k}) \, \text{m}
\]

### Step 1: Calculate the position vector at \( t = 1.00 \, \text{s} \)
\[
\vec{r}(1.00) = (9.60(1.00) \, \hat{i} + 8.85 \, \hat{j} - 1.00(1.00)^2 \, \hat{k}) \, \text{m}
\]
\[
\vec{r}(1.00) = (9.60 \, \hat{i} + 8.85 \, \hat{j} - 1.00 \, \hat{k}) \, \text{m}
\]

### Step 2: Calculate the position vector at \( t = 3.00 \, \text{s} \)
\[
\vec{r}(3.00) = (9.60(3.00) \, \hat{i} + 8.85 \, \hat{j} - 1.00(3.00)^2 \, \hat{k}) \, \text{m}
\]
\[
\vec{r}(3.00) = (28.80 \, \hat{i} + 8.85 \, \hat{j} - 9.00 \, \hat{k}) \, \text{m}
\]

### Step 3: Compute the displacement vector \( \Delta \vec{r} \)
The displacement vector is given by:
\[
\Delta \vec{r} = \vec{r}(3.00) - \vec{r}(1.00)
\]
\[
\Delta \vec{r} = (28.80 \, \hat{i} + 8.85 \, \hat{j} - 9.00 \, \hat{k}) - (9.60 \, \hat{i} + 8.85 \, \hat{j} - 1.00 \, \hat{k})
\]
\[
\Delta \vec{r} = (28.80 - 9.60) \, \hat{i} + (8.85 - 8.85) \, \hat{j} + (-9.00 + 1.00) \, \hat{k}
\]
\[
\Delta \vec{r} = (19.20 \, \hat{i} + 0 \, \hat{j} - 8.00 \, \hat{k}) \, \text{m}
\]

### Step 4: Calculate the average velocity \( \vec{v}_{\text{avg}} \)
The average velocity is given by:
\[
\vec{v}_{\text{avg}} = \frac{\Delta \vec{r}}{\Delta t}
\]
where \( \Delta t = 3.00 \, \text{s} - 1.00 \, \text{s} = 2.00 \, \text{s} \).

\[
\vec{v}_{\text{avg}} = \frac{(19.20 \, \hat{i} + 0 \, \hat{j} - 8.00 \, \hat{k})}{2.00} \, \text{m/s}
\]
\[
\vec{v}_{\text{avg}} = (9.60 \, \hat{i} + 0 \, \hat{j} - 4.00 \, \hat{k}) \, \text{m/s}
\]

### Final Answer
The average velocity of the particle between \( t = 1.00 \, \text{s} \) and \( t = 3.00 \, \text{s} \) is:
\[
\vec{v}_{\text{avg}} = 9.60 \, \hat{i} - 4.00 \, \hat{k} \, \text{m/s}
\]

Would you like further details or have any questions about this problem? Here are some related questions:

1. How would you determine the instantaneous velocity at \( t = 2.00 \, \text{s} \)?
2. What would be the particle's displacement vector from \( t = 0 \, \text{s} \) to \( t = 4.00 \, \text{s} \)?
3. How does the position vector change if the coefficient of \( t^2 \) in the \( \hat{k} \) direction were doubled?
4. What is the significance of the \( \hat{j} \) component being constant in this problem?
5. How would the average speed differ from the average velocity in this case?

**Tip:** When dealing with vectors, remember that each component of the vector should be treated independently in calculations like addition, subtraction, and differentiation.

The position of a particular particle as a function of time is given by r⃗ = (9.60t⋅i^ + 8.85j^ - 1.00t^2⋅k^)m, where t is in seconds. What is the average velocity of the particle between t = 1.00 s and t = 3.00 s? Express your answer in terms of the unit vectors i^, j^, and k^.

Discover how to calculate the average velocity of a particle given its position vector as a function of time. This problem involves key concepts in calculus and kinematics, such as displacement and average velocity. Step-by-step solutions provided for students at Grades 11-12 or college level.

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