AP Calculus AB Question 116: Answer And Explanation - Crack AP

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Question: 116

6. The maximum velocity attained on the interval 0 ≤ t ≤ 5, by the particle whose displacement is given by s(t) = 2t3 - 12t2 + 16t + 2 is

• A. 286
• B. 46
• C. 16
• D. 0

Correct Answer: B

Explanation:

B The maximum velocity attained on the interval 0 ≤ t ≤ 5 by the particle whose displacement is given by s(t) = 2t3 12t2 + 16t + 2 is

Velocity is the first derivative of position with respect to time.

The first derivative is

v(t) = 6t2 − 24t + 16

If we want to find the maximum velocity, we take the derivative of velocity (which is acceleration) and find where the derivative is zero.

v′(t) = 12t − 24

Next, we set the derivative equal to zero and solve for t, in order to find the critical value.

12t − 24 = 0

t = 2

Note that the second derivative of velocity is 12, which is positive. Remember the second derivative test: If the sign of the second derivative at a critical value is positive, then the curve has a local minimum there. If the sign of the second derivative is negative, then the curve has a local maximum there.

Thus, the velocity is a minimum at t = 2. In order to find where it has an absolute maximum, we plug the endpoints of the interval into the original equation for velocity, and the larger value will be the answer.

At t = 0 the velocity is 16. At t = 5, the velocity is 46.

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