# Finding Velocity and Position with a Given Acceleration Function

In this article, we will explore the concepts of velocity and position in the context of a given acceleration function and initial conditions. We will use vector calculus to derive the expressions for velocity and position, and then calculate the specific values based on the provided information.

## Problem Statement

Given the following acceleration function: $\mathbf{a}(t) = \langle 3e^t, 18t, 2e^{-t} \rangle$

and the initial conditions: $\mathbf{v}(0) = (3, 0, -6)$ $\mathbf{r}(0) = (6, -1, 2)$

Find the velocity and position vectors as functions of time.

## Solution

To find the velocity and position vectors, we need to integrate the acceleration function with respect to time, using the given initial conditions.

### Step 1: Find the Velocity Vector

The velocity vector is the derivative of the position vector with respect to time: $\mathbf{v}(t) = \frac{d\mathbf{r}(t)}{dt}$

Integrating the acceleration function, we get: $\mathbf{v}(t) = \int \mathbf{a}(t) dt = \int \langle 3e^t, 18t, 2e^{-t} \rangle dt = \langle 3e^t, 9t^2, -2e^{-t} \rangle + \mathbf{C}$

Using the initial condition $\mathbf{v}(0) = (3, 0, -6)$, we can find the constant vector $\mathbf{C}$: $\mathbf{C} = (3, 0, -6)$

Therefore, the velocity vector is: $\mathbf{v}(t) = \langle 3e^t, 9t^2, -2e^{-t} \rangle + (3, 0, -6)$

### Step 2: Find the Position Vector

The position vector is the integral of the velocity vector with respect to time: $\mathbf{r}(t) = \int \mathbf{v}(t) dt$

Integrating the velocity vector, we get: $\mathbf{r}(t) = \int \langle 3e^t, 9t^2, -2e^{-t} \rangle + (3, 0, -6) dt = \langle 3e^t, 3t^3, -2e^{-t} \rangle + (3t, 0, -6t) + \mathbf{D}$

Using the initial condition $\mathbf{r}(0) = (6, -1, 2)$, we can find the constant vector $\mathbf{D}$: $\mathbf{D} = (6, -1, 2)$

Therefore, the position vector is: $\mathbf{r}(t) = \langle 3e^t, 3t^3, -2e^{-t} \rangle + (3t, 0, -6t) + (6, -1, 2)$

## Conclusion

In this article, we have derived the expressions for the velocity and position vectors given the acceleration function and the initial conditions. The velocity vector is $\mathbf{v}(t) = \langle 3e^t, 9t^2, -2e^{-t} \rangle + (3, 0, -6)$, and the position vector is $\mathbf{r}(t) = \langle 3e^t, 3t^3, -2e^{-t} \rangle + (3t, 0, -6t) + (6, -1, 2)$. These expressions can be used to calculate the velocity and position at any given time.