Question

Find r(t) and v(t) given acceleration
a(t)=⟨t,9⟩,
initial velocity
v(0)=⟨−1,3⟩,
and initial position
r(0)=⟨0,0⟩.

Answer 1

$a(t) = t\textbf{i} + 9\textbf{j} \\ \frac{\mathrm{d}v}{\mathrm{d}t} = t\textbf{i} + 9\textbf{j} \\ \textsf{Separating variables, we have,}\\ \mathrm{d}v = (t\textbf{i} + 9\textbf{j})\mathrm{d}t\\ \int \mathrm{d}v = \int(t\textbf{i} + 9\textbf{j})\mathrm{d}t\\ v = \int t \mathrm{d}t\textbf{i} + \int9\mathrm{d}t\textbf{j}\\ v(t) = \left(\frac{t^2}{2} + C_1\right)\textbf{i} + (9t + C_2)\textbf{j}\\ v(t) = \frac{t^2}{2}\textbf{i} + 9t\textbf{j} + (C_1\textbf{i} + C_2\textbf{j})\\ v(0) = C_1\textbf{i} + C_2\textbf{j}\\ -\textbf{i} + 3\textbf{j} = C_1\textbf{i} + C_2\textbf{j}\\ \Rightarrow C_1 = -1, C_2=3\\ \therefore v(t) = \frac{t^2}{2}\textbf{i} + 9t\textbf{j} - \textbf{i} + 3\textbf{j}\\ v(t) = \left(\frac{t^2}{2} - 1\right)\textbf{i} + (9t + 3)\textbf{j}\\ \frac{\mathrm{d}r}{\mathrm{d}t} = \left(\frac{t^2}{2} - 1\right)\textbf{i} + (9t + 3)\textbf{j}\\ \mathrm{d}r = \left(\frac{t^2}{2} - 1\right)\textbf{i} \mathrm{d}t+(9t + 3)\textbf{j}\mathrm{d}t\\ \int\mathrm{d}r = \int\left(\frac{t^2}{2} - 1\right)\mathrm{d}t\textbf{i} + \int(9t + 3)\mathrm{d}t\textbf{j}\\ r(t) = \left(\frac{t^3}{6} - t + B_1\right)\textbf{i} + \left(\frac{9t^2}{2} + 3t + B_2\right)\textbf{j}\\ r(0) = B_1\textbf{i} + B_2\textbf{j}\\ 0\textbf{i} + 0\textbf{j}= B_1\textbf{i} + B_2\textbf{j}\\ \Rightarrow B_1 = B_2 = 0\\ \therefore r(t) = \left(\frac{t^3}{6} - t\right)\textbf{i} + \left(\frac{9t^2}{2} + 3t\right)\textbf{j} \\ \textsf{Writing} \hspace{0.1cm} v(t) \hspace{0.1cm} \textsf{and} \hspace{0.1cm} r(t) \\\textsf{as column vectors}\\ r(t) = \left(\frac{t^3}{6} - t, \frac{9t^2}{2} + 3t\right) \hspace{0.1cm} \&\\ v(t) = \left(\frac{t^2}{2} - 1,9t + 3\right)$