AP Calculus BC Question 62: Answer and Explanation
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7. A rock is thrown straight upward with an initial velocity of 50 m/s from a point 100 m above the ground. If the acceleration of the rock at any time t is a = -10 m/s2, what is the maximum height of the rock (in meters)?
- A. 150
- B. 175
- C. 200
- D. 225
Correct Answer: D
D Because the derivative of velocity with respect to time is acceleration, we have
v(t) = - 10 dt = - 10t + C
Now we can plug in the initial condition to solve for the constant.
50 = -10(0) + C
C = 50
Therefore, the velocity function is v(t) = -10t + 50.
Note that the velocity is zero at t = 5.
Next, because the derivative of position with respect to time is velocity, we have
s(t) = (- 10t + 50) dt = - 5t2 + 50t + C
Now, we can plug in the initial condition to solve for the constant.
100 = -5(0)2 + 50(0) + C
C = 100
Therefore, the position function is s(t) = -5t2 + 50t + 100.
The maximum height occurs when the velocity is zero, so we plug t = 5 into the position function to get s(5) = -5(5)2 + 50(5) + 100 = 225.