So there is a 1500 kg car travelling at 60km/h over a circular hill that has a vertical radius of 200 m. So the coefficient of static friction is 0.7 and the car starts applying the brakes when its 50 m out from the crest of the circular hill and the car is supposed to stop exactly on the crest. With this information I need to find the minimum stopping distance, which I believe I can get from finding the deceleration of the car and then plugging it into a kinematics equation.
Using some known physics equations I managed to find an equation that describes the acceleration of the car. It is:
$$a = 6.86 \cos \frac{(at^3)}{200} - 0.0035a^2t^2$$
Now I need to find the average acceleration between 0 and 2.43 s, for which I need to integrate the integral as such:
$$\int_{0}^{2.43} \left( 6.86 \cos \frac{(at^3)}{200} - 0.0035a^2t^2\right)\mathrm dt$$
So I'm just wondering is this even possible to do? Any kind of help will be greatly appreciated.
As @AntonVrdoljak commented, if your equation is correct, you cannot continue if you do not know at least an approximate function $a(t)$.
Rewriting your equation with whole numbers, we have $$a+\frac{7 a^2 t^2}{2000}-\frac{343}{50} \cos \left(\frac{a t^3}{200}\right)=0$$
Suppose that the argument of the cosine is small and use Taylor expansion around $a=0$ followed by power series reversion. This should give you things such as $$a=\frac{343}{50}-\frac{823543 t^2}{5000000}+\frac{1977326743 t^4}{250000000000}-\frac{40353607 t^6}{10000000000}+\cdots$$ and $$ \cos \left(\frac{at^3}{200}\right)=1-\frac{117649 t^6}{200000000}+\frac{282475249 t^8}{10000000000000}+\cdots$$
Now, you can compute the definite integral.