Magnitude of velocity 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 $524m$.

Asked this question earlier and while the answerer was extremely helpful the answer he gave me was incorrect or at least didn't give the answer my tutor is looking for. (I don't mean to appear ungrateful for the attempted help in my earlier question the gentleman was extremely helpful and nice.)

Help please
Hint: The acceleration would be the vector sum of the tangential and radial one, i.e $$|a|=\sqrt{(\frac{v^2}{r})^2+(\frac{dv}{dt})^2}$$
Now you have $a(t)=0.06t$, can you find $v(t)$? and subsequently $x(t)$? then you know $x=524m$, find t from this and then find v at that t and find |a| by the above relation.
Notice that the circular path of the car does not affect the velocity function. We use the radius only to find the distance traveled by the car so that we may use basic calculus to find the velocity and acceleration.

The distance is, as you have found, to be $\frac{1}{3} \times 2\pi(250),$ which is about $523.6.$

We know that
$$\int_{0}^{t} (0.06t + 5) dt = 523.6$$
$$\left[0.03t^{2} + 5t\right]_{0}^{t} = 523.6$$
$$0.03t^{3} + 5t^{2} = 523.6,$$
and we conclude that $t$ is about $9.94\text{s}.$

We can find the velocity simply by plugging in, so
$$v = 0.06 \times 9.94$$
$$v = \boxed{0.60\frac{\text{m}}{\text{s}}}.$$

Differentiating the velocity function, we have
$$a = \boxed{0.06\frac{\text{m}}{\text{s}^{2}}}.$$