Final answer:
To find the position vector of a particle with the given acceleration and initial velocity and position, we need to integrate the acceleration function twice. First, we find the velocity function by integrating the acceleration function, and then we find the position function by integrating the velocity function. Using the initial conditions, we can solve for the integration constants to obtain the specific position vector function.
The final position of the particle is:
\[ x(t) = \frac{11}{6} t^3 \]
\[ y(t) = e^t + t + 1 \]
\[ z(t) = e^{-t} + 1 \]
Step-by-step explanation:
To find the position vector \( \mathbf{r}(t) \) of a particle with the given acceleration \( \mathbf{a}(t) \), initial velocity \( \mathbf{v}(0) \), and initial position \( \mathbf{r}(0) \), you can follow these steps:
Given \( \mathbf{a}(t) = 11t \mathbf{i} \, e^t \mathbf{j} \, e^{-t} \mathbf{k} \), integrate each component separately with respect to time to find the velocity components.
1. **Integration of Acceleration to Find Velocity:**
\[ v_x(t) = \int 11t \, dt = \frac{11}{2} t^2 + C_1 \]
\[ v_y(t) = \int e^t \, dt = e^t + C_2 \]
\[ v_z(t) = \int e^{-t} \, dt = -e^{-t} + C_3 \]
2. **Use Initial Velocity:**
Apply \( \mathbf{v}(0) = \mathbf{k} \):
\[ v_x(0) = C_1 = 0 \]
\[ v_y(0) = C_2 = 1 \]
\[ v_z(0) = C_3 = 0 \]
So, we have:
\[ v_x(t) = \frac{11}{2} t^2 \]
\[ v_y(t) = e^t + 1 \]
\[ v_z(t) = -e^{-t} \]
3. **Integration of Velocity to Find Position:**
\[ x(t) = \int v_x(t) \, dt = \frac{11}{6} t^3 + C_4 \]
\[ y(t) = \int v_y(t) \, dt = e^t + t + C_5 \]
\[ z(t) = \int v_z(t) \, dt = e^{-t} + C_6 \]
4. **Use Initial Position:**
Apply \( \mathbf{r}(0) = \mathbf{j} \mathbf{k} \):
\[ x(0) = C_4 = 0 \]
\[ y(0) = C_5 = 1 \]
\[ z(0) = C_6 = 1 \]
So, the final position of the particle is:
\[ x(t) = \frac{11}{6} t^3 \]
\[ y(t) = e^t + t + 1 \]
\[ z(t) = e^{-t} + 1 \]