A car travels around a circular track with a radius of $r=250m$. When it is at point $A$ then $V_a=5m/s$ which increases at a rate of $\dot{v}=(0.06t)m/s$. Determine the magnitude of its velocity and its acceleration when it is $1/3$ around the track. My distance in this situation is obviously $524m$.
So I know I need to integrate the $0.06t$ which would give me $0.03t^2+5$. I am unsure where to go after this. Do I make $a=0$ and solve t for that? Or do I integrate further in which case I'd get $524=0.01t^3+5t+c$ (because of it being $1/3$ around the track) which i'd be unable to solve, or am I completely wrong?
Help please.
Why is that? That is the correct approach. You can just set $c=0$, as you are not interested in anything absolute (that is, you can just choose your coordinate system such that you begin "counting" the distance the car has traveled from $t=0$). Then use this time $t_{1/3}$ to solve for $v$ and $\dot v$ at that time.
EDIT (added the calculation):
$$\frac{2 \pi \cdot 250}{3}=0.01t^3+5t \leftrightarrow t\approx\{-589,88.9\}$$but we throw away the negative solution, since only the positive time makes sense physically, so $t_{1/3}=88.9$.
The we put this into our equations for $v$ and $\dot v$: $$\dot v(t_{1/3})=0.06 \cdot 88.9 = 5.33 \; m/s^2$$ and $$v(t_{1/3})=0.03 \cdot 88.9^2 +5=242.1 \; m/s$$