Let's go through each part of the question one by one.

Given:

• $x = 4t - 3$
• $y = 3t^2 + 5$
• $z = 7$

where $x$, $y$, and $z$ are in meters and $t$ is in seconds.

Part (a): Find the position vector as a function of time.

The position vector $\vec{r}(t)$ can be written in terms of $x$, $y$, and $z$ as: $\vec{r}(t) = x\hat{i} + y\hat{j} + z\hat{k}$

Substitute the expressions for $x$, $y$, and $z$: $\vec{r}(t) = (4t - 3)\hat{i} + (3t^2 + 5)\hat{j} + 7\hat{k}$

So, the position vector as a function of time is: $\vec{r}(t) = (4t - 3)\hat{i} + (3t^2 + 5)\hat{j} + 7\hat{k}$

Part (b): What is the initial position?

The initial position is found by substituting $t = 0$ into $\vec{r}(t)$.

$x(0) = 4(0) - 3 = -3$ $y(0) = 3(0)^2 + 5 = 5$ $z(0) = 7$

Thus, the initial position vector is: $\vec{r}(0) = -3\hat{i} + 5\hat{j} + 7\hat{k}$

Part (c): Find the displacement between $t = 1 \, \text{s}$ and $t = 2 \, \text{s}$.

Displacement vector, $\Delta \vec{r}$, is given by: $\Delta \vec{r} = \vec{r}(2) - \vec{r}(1)$

1. Calculate $\vec{r}(1)$:

   $x(1) = 4(1) - 3 = 1$ $y(1) = 3(1)^2 + 5 = 3 + 5 = 8$ $z(1) = 7$

   So, $\vec{r}(1) = 1\hat{i} + 8\hat{j} + 7\hat{k}$.

2. Calculate $\vec{r}(2)$:

   $x(2) = 4(2) - 3 = 8 - 3 = 5$ $y(2) = 3(2)^2 + 5 = 3(4) + 5 = 12 + 5 = 17$ $z(2) = 7$

   So, $\vec{r}(2) = 5\hat{i} + 17\hat{j} + 7\hat{k}$.

3. Calculate the displacement:

   $\Delta \vec{r} = \vec{r}(2) - \vec{r}(1) = (5 - 1)\hat{i} + (17 - 8)\hat{j} + (7 - 7)\hat{k}$ $\Delta \vec{r} = 4\hat{i} + 9\hat{j} + 0\hat{k}$

   Thus, the displacement vector between $t = 1 \, \text{s}$ and $t = 2 \, \text{s}$ is: $\Delta \vec{r} = 4\hat{i} + 9\hat{j}$

Part (d): Find the magnitude of the displacement vector between $t = 1 \, \text{s}$ and $t = 2 \, \text{s}$.

The magnitude of the displacement vector $\Delta \vec{r}$ is calculated as: $|\Delta \vec{r}| = \sqrt{(4)^2 + (9)^2 + (0)^2}$ $|\Delta \vec{r}| = \sqrt{16 + 81}$ $|\Delta \vec{r}| = \sqrt{97}$ $|\Delta \vec{r}| \approx 9.8 \, \text{meters}$

So, the magnitude of the displacement vector between $t = 1 \, \text{s}$ and $t = 2 \, \text{s}$ is approximately $9.8$ meters.

Would you like more details or have any questions?

Further Questions:

1. How do we calculate the velocity vector from the position vector?
2. How can we determine the acceleration vector for this motion?
3. What is the significance of the displacement vector being different from the position vector?
4. How would we find the total distance traveled instead of displacement?
5. Can we analyze this motion in terms of projectile motion principles?

Tip:

Remember, displacement is a vector quantity that considers only the initial and final positions, while distance considers the entire path traveled.