Problem Three: 1998-1999
A twenty pound bag full of severed thumbs is
dropped from the second highest floor of the Sears Tower, 1542 feet above
the ground. The top floor was rented out for a Narcoleptic Rabbi Association
(NRA) convention.
(a) Find the position and velocity functions
for dem thumbs.
(b) Determine the average sack velocity
on the interval [2,3].
(c) Find the velocity at t =
1, 2, 3, and 4 seconds.
(d) How long exactly (to the thousandths
place) does it take for dem thumbs to smash into the pavement below in
a disgusting, bloody mess?
(e) If the velocity of a bag of thumbs
required to kill a passer-by (fatal velocity of the thumb bag) is 250 ft/sec,
would a 6 foot tall man be killed on impact?
(f) If the "Becker Deadness Quotient,"
(c) 1998, is determined by the function
= Deadness
in deadograms (dgs),
exactly how dead is he?
(g) Wouldn't it be freaky if the guy
who got killed had cut off his own thumb because he thought it was talking
to him and threatening to kill him, and some thumb collector picked it
up and put it in this bag and now it finally did?!?!?
Solution:
(a) position function:
s(t) = -16t2 + 1542
velocity function:
v(t) = -32t
(b) Remember that the average velocity is given by the
slope of the secant line, rather than the slope of the tangent line. Plug 2 and
3 into s(t) to come up with two ordered pair, and find the slope of that
segment. The points are (2, 1398) and (3, 1478). Just as the tangent line is based
on the position function, the secant graph's points come from the position function.
The slope is (1478-1398) / (3-2) or 80 ft/sec. Don't forget units!
(c) Plug 1, 2, 3, and 4 into the velocity
function from part (a) to get:
v(1) = -32 ft/sec
v(2) = -64 ft/sec
v(3) = -96 ft/sec
v(4) = -128 ft/sec
(d) We want to know when the position of the thumb bag (relative
to the ground) is 0 ft (i.e. on the ground). So, we set the position equation
equal to zero: -16t2 + 1542 = 0. Now, we solve for t by subtracting
the constant from both sides: -16t2 = -1542. Divide by -16 to get t2
= 96.375. Take the square root of both sides to get t = 9.817 seconds.
Don't forget that the AP test wants answers rounded or truncated to the thousandths
place.
(e) We want to find the velocity of the bag when it is six feet
off the ground. First, we need to find out at what time the bag reaches that speed
by setting the position equation equal to 6 (the top of the unfortunate man's
head). We get -16t2 + 1542 = 6. Solve for t like in part
(d) to get t = 9.798 seconds. Now, plug that value of t into the velocity
equation to ascertain the velocity at that point: v(9.798) = -32(9.798)
= -313.535 ft/sec.
(f) Plug into the Becker Deadness Quotient
to get D = 6.354 dgs.
(g) Yes. Yes it would.
Special Celebrity
Guest Answer to part (g) by Stuart: "No. It would be fascinating."
