# How far can the car travel before it loses its grip? Why is my solution wrong?

A car is traveling on a circular race track with radius $$R=100m$$ with the constant acceleration $$a_0 = 0.3g.$$ The friction constant for the road and tires is $$\mu = 0.5.$$ I'm asked to determine how far the car can travel before it loses its grip.

My attempt:

I know that friction causes the car to turn and using the natural coordinate system, $$a = \ddot{s}e_t + \frac{\dot{s}^2}{R}e_n$$ I realize that for the car to maintain its path the "turning acceleration", $$\frac{\dot{s}^2}{R}$$ cannot exceed the maximal frictional force $$0.5mg.$$ So,

$$0.5mg = \frac{\dot{s}^2}{R} \implies \dot{s} =\sqrt{R0.5mg}.$$ Now I need to find out after what time this speed is aquired, $$0.3gt = \sqrt{R0.5mg},$$ $$t = \frac{\sqrt{R0.5mg}}{0.3g}$$ The distance traveled after $$t$$ is $$\frac{0.3gt^2}{2} = \frac{25m}{0.3}.$$ However this answer is wrong, it is supposed to be $$67$$ meters. (I even have mass in my answer, which shouldnt be there) Why is my solution wrong?

You have to include both acceleration: The radial one and the azimuthal. Using Pythagoras we obtain the equation $$a_{rad}^2 + a_{az}^2 = a_{max}^2$$ where
$$a_{rad} = v^2/R = (a t)^2/R$$, $$a_{az} = a$$, and $$a_{max} = \mu g$$. Solve this equation for $$t^2$$. This yields the correct result!