We are given the following information:

• Period $T = 2.0 \, \text{s}$
• Radius $r = 2.60 \, \text{m}$
• Acceleration vector $\vec{a} = (6.80 \, \text{m/s}^2) \hat{i} + (-3.80 \, \text{m/s}^2) \hat{j}$

We need to find:

1. $\vec{v} \cdot \vec{a}$ (dot product of velocity and acceleration vectors)
2. $\vec{r} \times \vec{a}$ (cross product of position and acceleration vectors)

Step 1: Find $\vec{v} \cdot \vec{a}$

The dot product of velocity and acceleration, $\vec{v} \cdot \vec{a}$, is zero for uniform circular motion. This is because in uniform circular motion, the velocity vector $\vec{v}$ is always perpendicular to the acceleration vector $\vec{a}$ (which points toward the center of the circle).

Thus, $\vec{v} \cdot \vec{a} = 0$

Step 2: Find $\vec{r} \times \vec{a}$

The cross product $\vec{r} \times \vec{a}$ involves finding the cross product of the position vector and the acceleration vector. At $t_1$, we assume the position vector $\vec{r}$ is given by:

$\vec{r} = (2.60 \, \text{m}) \hat{i}$

Now, calculate the cross product:

$\vec{r} \times \vec{a} = (2.60 \, \hat{i}) \times \left( (6.80 \, \hat{i}) + (-3.80 \, \hat{j}) \right)$

Using the properties of the cross product:

$\hat{i} \times \hat{i} = 0 \quad \text{and} \quad \hat{i} \times \hat{j} = \hat{k}$

Thus, the cross product simplifies to:

$\vec{r} \times \vec{a} = 2.60 \, \hat{i} \times (-3.80 \, \hat{j}) = -9.88 \, \hat{k} \, \text{m}^3/\text{s}^2$

Final Answers:

(a) $\vec{v} \cdot \vec{a} = 0 \, \text{m}^2/\text{s}^3$

(b) $\vec{r} \times \vec{a} = -9.88 \, \hat{k} \, \text{m}^3/\text{s}^2$

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Would you like more details on any step or concept here? Let me know if you have any questions!

Here are 5 related questions to expand your understanding:

1. How does uniform circular motion affect the velocity and acceleration vectors?
2. What is the physical significance of the dot product being zero in uniform circular motion?
3. Why is the acceleration vector directed toward the center in uniform circular motion?
4. Can the radius of circular motion affect the magnitude of centripetal acceleration?
5. How would the cross product $\vec{r} \times \vec{v}$ differ from $\vec{r} \times \vec{a}$?

Tip: In circular motion, centripetal acceleration always points toward the center of the circle and is responsible for constantly changing the direction of the velocity.