Problem Six: 1999-2000

Calculus I (AB): A Review of Limits and Continuity

Graph the function g that satisfies all of the following conditions:
        

Difficulty Rating: 
Solution:
This graph is just one solution...there may be others possible:

Calculus II (BC): A Review of Derivatives

During a spectacular stunt (and a slight weapons malfunction), Mr. Kelley's torso is hurled into the air. The position of his body (in feet) as it sails through the air at time t seconds is given by:

h(t) = -16t2 + 38t + 15.

    (a)    What is height of the launching platform on the hill from which Kelley is shot?
    (b)    What is the velocity of his spinning torso at t = 0.34 seconds?
    (c)    Say the goal of the "experiment" was to make Kelley the first man ever launched over the world's largest tree (The Tree of Doom). If this tree is 45 feet tall (we must also assume we've gone to a world where we have really small trees), is he able to clear the tree? If he did, by how much did he clear it, and if he didn't, how far from the top was he?
    (d)    On his way down (assuming the tree didn't slow him down), Kelley runs flat into a brick building. For him to fly through the wall and leave a Kelley-shaped hole (like in cartoons), he must have a velocity of exactly -43.6 ft/sec. Faster and he shatters--slower and he bounces. At what height does he travel this velocity and break through the walls of the Hospital for the Criminally Insane? (Ohhh...hello Doctor Lecter ...)

Difficulty Rating: 


Solution:
(a)    Original height is where t = 0, so plug in 0 for t and get h(0) = 15 feet.
(b)    Because velocity is the derivative (rate of change) of position, we find the derivative of h and call it v(t), as h'(t) = v(t):
v(t) = -32t +38
Now, find v(.34) to get that the velocity is 27.12 ft/sec.
(c)    The maximum height of h is where its derivative will be zero (or undefined, but there is no such spot on this curve). Velocity will be zero at t = 1.1875 seconds. Let's see what height he had at that time in order to see if it was high enough to top the tree. To do that, we plug t = 1.1875 (where the maximum occurs) into the position function, and we get h(1.1875) = 37.5625 feet. Guess what? He was 45-37.5625 = 7.438 feet too short. So close, and yet so far away ... and he thought the branch of his local bank was tough ...
(d)    When is the velocity equal to -43.6 ft/sec? Set v(t) equal to that number and solve for t. It turns out that at time = 2.55 sec, Kelley was traveling that speed. So, his height at that time was h(2.55) = 7.86 feet, which has to be the ground floor--he got lucky.


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