The acceleration, in meters per second per second, of a race car is modeled by A(t)=t3−152t2+12t+10, where t is measured in seconds. What is the car's maximum acceleration on the time interval 0≤t≤6 ?

  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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Additional comment

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].

the answer to the question is

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