I am going to assume, even though it is not very clear from the question, that the part of the track on which the car accelerates is straight. Then, the motion of the car on this part of the track can be described by the equation of the one-dimensional motion with constant acceleration:

`d = v_it + 1/2at^2` .

Here, d is the displacement (or in this case, the length of the track on which the car accelerates), v_i is the initial velocity, a is the acceleration and t is the time during which the car accelerates. All these quantities, except for the acceleration, which we have to find, are given in the question.

`2.20 cm = 29.0 (cm)/s*0.0376 s + 1/2 a (0.0376 s)^2` .

Solving for a, we get

`a = ((2.20 - 29.0*0.0376)*2)/(0.0376)^2 (cm)/s^2 =1570 (cm)/s^2 = 1.57*10^3 (cm)/s^2`

The acceleration the track provides to the car is 1570 cm/s^2, or 15.7 m/s^2.