We are given the acceleration of the rocket as a vector function:

$\vec{a}(t) = \langle \cos(t), 4, e^{-t} \rangle$

The position at time $t = 0$ is:

$\vec{r}(0) = \langle 0, 0, 0 \rangle$

And the velocity at time $t = 0$ is:

$\vec{v}(0) = \langle 1, 2, 3 \rangle$

We need to find the position at $t = 2$. To solve this, we need to integrate the acceleration function twice to find the position function.

Step 1: Find the velocity

To get the velocity, we integrate the acceleration:

$\vec{v}(t) = \int \vec{a}(t) \, dt = \int \langle \cos(t), 4, e^{-t} \rangle \, dt$

This gives:

$\vec{v}(t) = \langle \sin(t), 4t, -e^{-t} \rangle + \vec{C}$

To find $\vec{C}$, we use the initial velocity at $t = 0$:

$\vec{v}(0) = \langle 1, 2, 3 \rangle$

Substitute $t = 0$ into the equation for $\vec{v}(t)$:

$\langle \sin(0), 4(0), -e^{-0} \rangle + \vec{C} = \langle 1, 2, 3 \rangle$

This simplifies to:

$\langle 0, 0, -1 \rangle + \vec{C} = \langle 1, 2, 3 \rangle$

Thus, the constant vector $\vec{C} = \langle 1, 2, 4 \rangle$.

So, the velocity function is:

$\vec{v}(t) = \langle \sin(t) + 1, 4t + 2, -e^{-t} + 4 \rangle$

Step 2: Find the position

Next, we integrate the velocity function to get the position:

$\vec{r}(t) = \int \vec{v}(t) \, dt = \int \langle \sin(t) + 1, 4t + 2, -e^{-t} + 4 \rangle \, dt$

This gives:

$\vec{r}(t) = \langle -\cos(t) + t + C_1, 2t^2 + 2t + C_2, e^{-t} + 4t + C_3 \rangle$

To find the constants $C_1$, $C_2$, and $C_3$, we use the initial condition $\vec{r}(0) = \langle 0, 0, 0 \rangle$:

For the first component:

$-\cos(0) + 0 + C_1 = 0 \implies C_1 = 1$

For the second component:

$2(0)^2 + 2(0) + C_2 = 0 \implies C_2 = 0$

For the third component:

$e^{0} + 4(0) + C_3 = 0 \implies C_3 = -1$

Thus, the position function is:

$\vec{r}(t) = \langle -\cos(t) + t + 1, 2t^2 + 2t, e^{-t} + 4t - 1 \rangle$

Step 3: Evaluate at $t = 2$

Substitute $t = 2$ into the position function:

$\vec{r}(2) = \langle -\cos(2) + 2 + 1, 2(2)^2 + 2(2), e^{-2} + 4(2) - 1 \rangle$

Simplifying each component:

1. $-\cos(2) + 2 + 1 = 3 - \cos(2)$
2. $2(4) + 2(2) = 8 + 4 = 12$
3. $e^{-2} + 8 - 1 = 7 + e^{-2}$

Thus, the position at $t = 2$ is:

$\vec{r}(2) = \langle 3 - \cos(2), 12, 7 + e^{-2} \rangle$

Final Answer:

The correct answer is A. $\langle 3 - \cos(2), 12, 7 + e^{-2} \rangle$.

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Would you like more details or have any further questions on this topic?

Here are 5 related questions to explore:

1. How do we calculate the acceleration from the velocity and position functions?
2. How do initial conditions affect the solution of differential equations in motion problems?
3. Can you describe how to handle vector integrals in more complex motion problems?
4. What does it mean physically when an acceleration has both time-dependent and constant components?
5. How would the answer change if the initial velocity or position were different?

Tip: To solve problems like this, remember to first integrate the acceleration to get the velocity, then integrate the velocity to find the position. Always use initial conditions to solve for constants of integration.