Given the position vector of the particle:

Given the position vector of the particle:

$$\vec{r}(t) = (\cos(t)\hat{i} + \sin(t)\hat{j}) \, \text{m}$$

Step 1: Find the velocity and acceleration vectors.
The velocity vector $\vec{v}(t)$ is the first derivative of the position vector with respect to time $t$:

$$\vec{v}(t) = \frac{d\vec{r}}{dt} = \frac{d}{dt}(\cos(t)\hat{i} + \sin(t)\hat{j})$$ $$\vec{v}(t) = (-\sin(t)\hat{i} + \cos(t)\hat{j}) \, \text{m/s}$$

The acceleration vector $\vec{a}(t)$ is the first derivative of the velocity vector with respect to time $t$ (or the second derivative of the position vector):

$$\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}(-\sin(t)\hat{i} + \cos(t)\hat{j})$$ $$\vec{a}(t) = (-\cos(t)\hat{i} - \sin(t)\hat{j}) \, \text{m/s}^2$$

a) Show that the position and acceleration vectors of the particle are in opposite directions.
To show that two vectors are in opposite directions, one must be a negative scalar multiple of the other.
We have $\vec{r}(t) = \cos(t)\hat{i} + \sin(t)\hat{j}$ and $\vec{a}(t) = -\cos(t)\hat{i} - \sin(t)\hat{j}$.
We can factor out $-1$ from $\vec{a}(t)$:

$$\vec{a}(t) = -(\cos(t)\hat{i} + \sin(t)\hat{j})$$

Comparing this with $\vec{r}(t)$, we see that:

$$\vec{a}(t) = -\vec{r}(t)$$

Since the acceleration vector is equal to the negative of the position vector, they are always in opposite directions.

b) Show that the velocity and acceleration vectors are always at right angles to each other.
Two vectors are at right angles (perpendicular) if their dot product is zero.
We have $\vec{v}(t) = -\sin(t)\hat{i} + \cos(t)\hat{j}$ and $\vec{a}(t) = -\cos(t)\hat{i} - \sin(t)\hat{j}$.
Calculate their dot product:

$$\vec{v}(t) \cdot \vec{a}(t) = (-\sin(t))(-\cos(t)) + (\cos(t))(-\sin(t))$$ $$\vec{v}(t) \cdot \vec{a}(t) = \sin(t)\cos(t) - \cos(t)\sin(t)$$ $$\vec{v}(t) \cdot \vec{a}(t) = 0$$

Since the dot product of the velocity and acceleration vectors is always zero, they are always at right angles to each other.

c) Given $t = \pi$, find the position vector.
Substitute $t = \pi$ into the position vector equation:

$$\vec{r}(\pi) = \cos(\pi)\hat{i} + \sin(\pi)\hat{j}$$

We know that $\cos(\pi) = -1$ and $\sin(\pi) = 0$.

$$\vec{r}(\pi) = (-1)\hat{i} + (0)\hat{j}$$ $$\boxed{\vec{r}(\pi) = -\hat{i} \, \text{m}}$$

d) This part of the question is incomplete. However, the given information for two particles P and Q is:

• Initial position vector of particle P: $\vec{r}_P = (3\hat{i} + 2\hat{j}) \, \text{m}$
• Initial position vector of particle Q: $\vec{r}_Q = (-2\hat{i} + 5\hat{j}) \, \text{m}$
• Constant velocity of particle P: $\vec{v}_P = (-2\hat{i} - 5\hat{j}) \, \text{m/s}$
• Constant velocity of particle Q: $\vec{v}_Q = (3\hat{i} + 5\hat{j}) \, \text{m/s}$