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.