Question

Find r(t) if:
a(t) = 2i + 12tj
v(0) = 7i
r(0) = 2i + 9k

Answer 1

Note this is an initial value problem.

Recall that \(\mathbf{a}(t)=\mathbf{v}'(t)=\mathbf{r}''(t),\mathbf{v}(t)=\mathbf{r}'(t)\). So given \(\mathbf{a}(t)\), we should be able to find its antiderivative to determine \(\mathbf{v}(t)\), right? Let's try that out:$$\mathbf{a}(t)=2\mathbf{i}+12t\mathbf{j}\\\mathbf{v}(t)=\int\left(2\mathbf{i}+12t\mathbf{j}\right)\,\mathrm{d}t=2t\mathbf{i}+6t^2\mathbf{j}$$Finished, right? ... not quite. We still have a constant of integration, in this context a constant *vector* of integration \(\mathbf{C}\):$$\mathbf{v}(t)=2t\mathbf{i}+6t^2\mathbf{j}+\mathbf{C}$$Well, how do we figure out \(\mathbf{C}\)? Well, like any other IVP, we use our ICs (initial conditions) -- here, we're given \(\mathbf{v}(0)=7\mathbf{i}\). What happens if we plug in \(0\) into our antiderivative? $$\mathbf{v}(0)=2(0)\mathbf{i}+6(0)^2\mathbf{j}+\mathbf{C}=\mathbf{C}$$... in other worse, we can conclude \(\mathbf{C}=7\mathbf{i}\). Put it all together now:$$\mathbf{v}(t)=2t\mathbf{i}+6t^2\mathbf{j}+7\mathbf{i}=(2t+7)\mathbf{i}+6t^2\mathbf{j}$$

Answer 2

The second half of our problem is nearly identical, except now we're going to note \(\mathbf{v}(t)=\mathbf{r}'(t)\) -- so we can take its antiderivative to determine our position function. Once again, we'll end up with a constant of integration, so use your initial condition of \(\mathbf{r}(0)=2\mathbf{i}+9\mathbf{k}\) to finish the problem off.

Answer 3

@oldrin.bataku