Find the acceleration of the specified object. (Hint: Recall that if a variable is changing at a constant rate, its acceleration is zero.)
A boat is pulled into a dock by means of a winch 24 feet above the deck of the boat (see figure). The winch pulls in rope at a rate of 5 feet per second. Find the acceleration (in ft/sec2) of the boat when there is a total of 25 feet of rope out. (Round your answer to three decimal places.)
If the distance from the dock is x, then the length of rope z is related by
z^2 = x^2 + 24^2
when z=25, x=7 (the familiar 7-24-25 right triangle)
so, taking derivatives with respect to t, we have
zz' = xx'
z' + zz" = x' + xx"
We need to find x' to find x" (z"=0 since z' = -5 is constant)
25(-5) = 7x'
x' = -125/7 ft/s
Now we have
-5 = -125/7 + 7x"
x" = -90/7 ft/s^2
The negative sign indicates that the distance is decreasing -- the boat is accelerating toward the dock.
h^2 + x^2 = L^2
L = 25 - 5 t
h = 24
so
576 + x^2 = 625 -250 t + 25 t^2
x^2 = 49 - 250t + 25 t^2
2 x dx/dt = -250 + 50 t
x dx/dt = -125 + 25 t
x (d*2x/dt^2) + (dx/dt )^2 = 25
at t = 0, x = sqrt 49 = 7 and dx/dt = -250/14 = - 17.9
7 d^2x/dt^2 + 49 = 25
d^2 x/dt^2 = -24/7 = -3.43 ft/s^2