To find the position vector $\mathbf{r}(t)$ of the particle, we need to integrate the given acceleration vector $\mathbf{a}(t)$ twice and apply the initial conditions for velocity and position.

Step 1: Find the velocity vector $\mathbf{v}(t)$

The acceleration vector is given as: $\mathbf{a}(t) = 17t\,\hat{i} + e^t\,\hat{j} + e^{-t}\,\hat{k}$ We integrate $\mathbf{a}(t)$ to find the velocity vector $\mathbf{v}(t)$.

Velocity in the $\hat{i}$-direction:

$v_i(t) = \int 17t \, dt = \frac{17t^2}{2} + C_i$

Velocity in the $\hat{j}$-direction:

$v_j(t) = \int e^t \, dt = e^t + C_j$

Velocity in the $\hat{k}$-direction:

$v_k(t) = \int e^{-t} \, dt = -e^{-t} + C_k$

Thus, the general form of the velocity vector is: $\mathbf{v}(t) = \left( \frac{17t^2}{2} + C_i \right)\hat{i} + (e^t + C_j)\hat{j} + \left( -e^{-t} + C_k \right)\hat{k}$

Step 2: Apply the initial condition $\mathbf{v}(0) = \hat{k}$

From $\mathbf{v}(0) = \hat{k}$, we know: $\mathbf{v}(0) = C_i\,\hat{i} + (e^0 + C_j)\,\hat{j} + \left( -e^0 + C_k \right)\hat{k} = 0\,\hat{i} + 0\,\hat{j} + 1\,\hat{k}$

This gives the following conditions: $C_i = 0, \quad e^0 + C_j = 0 \implies C_j = -1, \quad -e^0 + C_k = 1 \implies C_k = 2$

Thus, the velocity vector becomes: $\mathbf{v}(t) = \frac{17t^2}{2}\,\hat{i} + (e^t - 1)\,\hat{j} + (-e^{-t} + 2)\,\hat{k}$

Step 3: Find the position vector $\mathbf{r}(t)$

We now integrate $\mathbf{v}(t)$ to find $\mathbf{r}(t)$.

Position in the $\hat{i}$-direction:

$r_i(t) = \int \frac{17t^2}{2} \, dt = \frac{17t^3}{6} + D_i$

Position in the $\hat{j}$-direction:

$r_j(t) = \int (e^t - 1) \, dt = e^t - t + D_j$

Position in the $\hat{k}$-direction:

$r_k(t) = \int (-e^{-t} + 2) \, dt = e^{-t} + 2t + D_k$

Thus, the general form of the position vector is: $\mathbf{r}(t) = \left( \frac{17t^3}{6} + D_i \right)\hat{i} + (e^t - t + D_j)\hat{j} + \left( e^{-t} + 2t + D_k \right)\hat{k}$

Step 4: Apply the initial condition $\mathbf{r}(0) = \hat{j} + \hat{k}$

From $\mathbf{r}(0) = \hat{j} + \hat{k}$, we know: $\mathbf{r}(0) = D_i\,\hat{i} + (e^0 - 0 + D_j)\,\hat{j} + \left( e^0 + 0 + D_k \right)\,\hat{k} = 0\,\hat{i} + 1\,\hat{j} + 1\,\hat{k}$

This gives the following conditions: $D_i = 0, \quad e^0 + D_j = 1 \implies D_j = 0, \quad e^0 + D_k = 1 \implies D_k = 0$

Thus, the position vector becomes: $\mathbf{r}(t) = \frac{17t^3}{6}\,\hat{i} + (e^t - t)\,\hat{j} + (e^{-t} + 2t)\,\hat{k}$

Final Answer:

The position vector of the particle is: $\mathbf{r}(t) = \frac{17t^3}{6}\,\hat{i} + (e^t - t)\,\hat{j} + (e^{-t} + 2t)\,\hat{k}$

Would you like further details on any part of the solution or have any questions?

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Here are 5 related questions:

1. How do you find velocity from a given acceleration vector?
2. What is the relationship between position, velocity, and acceleration vectors?
3. Can you derive the position vector if given initial conditions for position only?
4. How does the exponential function behave in differential equations?
5. What is the physical interpretation of the constants found during integration?

Tip: Always apply initial conditions immediately after integration to simplify the process of finding constants.