A particle moves along the circle $x^{2}+y^{2}=25$. Find its acceleration when it is at $(3,4)$.

Question

A particle moves along the circle $x^{2}+y^{2}=25$ at constant speed, making one revolution in $2$ $s$. Find its acceleration when it is at $(3,4)$.

What I got so far:

Let $x=5\cos(\pi t)$ and $y=5\sin(\pi t)$. Then $\mathbf{r}(t)=(5\cos(\pi t),5\sin(\pi t))$ with period $\frac{2\pi}{b}=\frac{2\pi}{\pi}=2$ $s$. Then I calculated $\mathbf{a}(t)=(-5\pi^{2} \cos(\pi t), -5\pi ^{2}\sin(\pi t))$.

What do I do next with the point $(3,4)$? I tried substituting it in $\mathbf{a}(t)$ but it doesn't seem like the correct thing to do. Do I use $\mathbf{r}(t)$? If so, does $t=(3,4)$?

Answer 1

Well, you could notice from here that $$ \boldsymbol{a}(t) = (−5\pi^2 \cos \pi t,−5\pi^2 \sin \pi t) = (-\pi^2) \cdot(5 \cos \pi t, 5 \sin \pi t) = -\pi^2 \boldsymbol{r}(t). $$

Now you don't have to calculate $t$, you can just plug in the $\boldsymbol{r}(t)$ and obtain $\boldsymbol{a}(t)$.

Answer 2

You have to find the $t$ that gives you $\mathbf{r}=(3,4)$, then plug that into your (correct) expression for $\mathbf{a}$.