A car starts from rest on a curve with a radius of $150m$ and tangential acceleration of $\displaystyle 1.5\frac{m}{s^2}$.
Through what angle will the car have traveled when the magnitude of its total acceleration is $\displaystyle 2.2\frac{m}{s^2}$?
A friend and I have been working on this for an hour and just cannot get the right answer.
We could use some help finding an equation that might be useful.
We used the pythagorean theorem to get the radial acceleration ($\displaystyle 1.6\frac{m}{s^2}$) and then used $\displaystyle A = \frac{v^2}{r}$ to get velocity (we got $\displaystyle 15.53 \frac{m}{s}$).
We don't know what to do past there.
Best Answer
You are simply trying to get the distance traveled around the curve in order to get the velocity. I think you are correct that radial acceleration is
$$a_r = \sqrt{a^2-a_t^2} = \sqrt{(2.2)^2-(1.5)^2} \frac{\text{m}}{\text{sec}^2} \approx 1.6 \frac{\text{m}}{\text{sec}^2} $$
and you know that
$$a_r = \frac{v^2}{r} \implies v = \sqrt{a_r r} \approx 15.5 \frac{\text{m}}{\text{sec}}$$
This speed corresponds to tangential velocity. The distance traveled around the curve is
$$d = \frac12 a_t t^2$$
where the time $t$ is known from the relation $v = a_t t$, which means that
$$d = \frac{v^2}{2 a_t} = \frac{a_r r}{2 a_t}$$
The angle through which the car has traveled is given by $\theta = d/r$, or
$$\theta = \frac{a_r}{2 a_t} = \frac12 \sqrt{\frac{a^2}{a_t^2}-1} \approx 0.536 \, \text{rad} \approx 30.7^{\circ}$$