Question

Given an acceleration vector, initial velocity $\left(u_{0}, v_{0}, w_{0}\right),\( and initial position \)\left\langle x_{0}, y_{0}, z_{0}\right\rangle,\( find the velocity and position vectors, for \)t \geq 0$. $$\begin{array}{l} \mathbf{a}(t)=\left\langle t, e^{-t}, 1\right\rangle,\left\langle u_{0}, v_{0}, w_{0}\right\rangle=\langle 0,0,1\rangle \\ \left\langle x_{0}, y_{0}, z_{0}\right\rangle=\langle 4,0,0\rangle \end{array}$$

Short answer

Answer: The velocity vector \(\mathbf{v}(t)\) is given by \(\left\langle \frac{1}{2}t^2, -e^{-t}, t+1\right\rangle\), and the position vector \(\mathbf{r}(t)\) is given by \(\left\langle \frac{1}{6}t^3+4, e^{-t}, \frac{1}{2}t^2+t\right\rangle\), for \(t \geq 0\).

Step-by-step solution

Find the velocity vector from the given acceleration vector

To find the velocity vector \(\mathbf{v}(t)\), integrate the acceleration vector component-wise with respect to time and add the initial velocity components. \(\mathbf{v}(t) = \int \mathbf{a}(t) dt + \mathbf{v}(0)\) \(v_x(t) = \int t dt + u_0\) \(v_y(t) = \int e^{-t} dt + v_0\) \(v_z(t) = \int 1 dt + w_0\)

Integrating and adding the initial velocity components

Integrate the acceleration components with respect to time, and add the initial velocity components. \(v_x(t) = \int t dt + u_0 = \frac{1}{2}t^2 + 0 = \frac{1}{2}t^2\) \(v_y(t) = \int e^{-t} dt + v_0 = -e^{-t} + 0 = -e^{-t}\) \(v_z(t) = \int 1 dt + w_0 = t + 1\) So, the velocity vector is \(\mathbf{v}(t) = \left\langle \frac{1}{2}t^2, -e^{-t}, t+1\right\rangle\).

Finding the position vector from the velocity vector

To find the position vector \(\mathbf{r}(t)\), integrate the velocity vector component-wise with respect to time, and add the initial position components. \(\mathbf{r}(t) =\int \mathbf{v}(t) dt +\mathbf{r}(0)\) \(x(t)= \int \frac{1}{2}t^2 dt + x_0\) \(y(t)= \int -e^{-t} dt+y_0\) \(z(t)= \int (t+1) dt + z_0\)

Integrating and adding the initial position components

Integrate the velocity components with respect to time and add the initial position components. \(x(t)= \int \frac{1}{2}t^2 dt + x_0 = \frac{1}{6}t^3 + 4\) \(y(t)= \int -e^{-t} dt+y_0 =e^{-t}+ 0= e^{-t}\) \(z(t)= \int (t+1) dt + z_0 = \frac{1}{2}t^2+t+0\) So, the position vector is \(\mathbf{r}(t) = \left\langle \frac{1}{6}t^3+4, e^{-t}, \frac{1}{2}t^2+t\right\rangle\). Thus, the velocity vector \(\mathbf{v}(t)\) is given by \(\left\langle \frac{1}{2}t^2, -e^{-t}, t+1\right\rangle\), and the position vector \(\mathbf{r}(t)\) is given by \(\left\langle \frac{1}{6}t^3+4, e^{-t}, \frac{1}{2}t^2+t\right\rangle\), for \(t \geq 0\).