Answer:
10.237 m/s²
Explanation:
The maximum acceleration will be at one end of the interval, or at a turning point in the interval. The turning points can be found where the derivative of acceleration is zero
A'(t) = 3t^2 -304t +12
The quadratic formula will tell the zeros of this.
t = (304±√(304^2 -4(3)(12)))/(2(3)) = (304 -√92272)/6 = (152 -√23068)/3
t ≈ 0.0394891
Then the maximum acceleration is ...
A(0.0394891) ≈ 10.23690 . . . . m/s^2
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<em>Additional comment</em>
The maximum magnitude of acceleration is at t=6, where acceleration is ...
A(6) = -5174 . . . . m/s^2
This is more than 500 g's in the negative direction, so the race car is not expected to survive that long.
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The attached graph shows the acceleration curve for positive acceleration. The acceleration continues to decrease to a minimum of -5174 m/s^2 at the right end of the interval [0, 6].
