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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?

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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!

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