91Ó°ÊÓ

We are given the acceleration vector \(\mathbf{a}(t)=\left\langle 1, t, 4t \right\rangle\), and to find the velocity vector, each component of the acceleration vector must be integrated with respect to time. This gives us: $$\mathbf{v}(t) = \left\langle\int 1 dt, \int t dt, \int 4t dt \right\rangle = \left\langle t + C_1, \frac{1}{2}t^2 + C_2, 2t^2 + C_3 \right\rangle$$.

Now, we need to integrate each component of the velocity vector with respect to time to find the position vector: $$\mathbf{r}(t) = \left\langle\int (t + C_1) dt, \int (\frac{1}{2}t^2 + C_2) dt, \int (2t^2 + C_3) dt \right\rangle = \left\langle \frac{1}{2}t^2 + C_1t + C_4, \frac{1}{6}t^3 + C_2t + C_5, \frac{2}{3}t^3 + C_3t + C_6 \right\rangle$$.

We are given the information that the initial position vector is \(\left\langle x_{0}, y_{0}, z_{0}\right\rangle =\left\langle 0, 0, 0\right\rangle\) and the initial velocity vector is \(\left\langle u_{0}, v_{0}, w_{0}\right\rangle =\left\langle 20, 0, 0\right\rangle\). We substitute these values into our equations for the velocity and position vectors when t = 0: For the initial velocity: \(\mathbf{v}(0) =\left\langle 20, 0, 0 \right\rangle = \left\langle 0 + C_1, 0 + C_2, 0 + C_3 \right\rangle\). So, \(C_1=20, C_2=0,\) and \(C_3=0\). For the initial position: \(\mathbf{r}(0) =\left\langle 0, 0, 0 \right\rangle = \left\langle 0 + 0 + C_4, 0 + 0 + C_5, 0 + 0 + C_6 \right\rangle\). So, \(C_4=0, C_5=0,\) and \(C_6=0\). Now, we can write the full velocity and position vectors, substituting the constants we have found: $$\mathbf{v}(t) = \left\langle 20+t, \frac{1}{2}t^2, 2t^2 \right\rangle$$ $$\mathbf{r}(t) = \left\langle \frac{1}{2}t^2 + 20t, \frac{1}{6}t^3, \frac{2}{3}t^3 \right\rangle$$ There you have it - the final forms for the velocity and position vectors, which are valid for \(t\geq 0\).