How fast is cart B moving toward Q at the instant when cart A is 5ft from Q?

Question

Two carts, A and B, are connected by a rope 39ft long that passes over a pulley P. The point Q is on the floor h = 12ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 1ft/s. How fast is cart B moving toward Q at the instant when cart A is 5ft from Q?

Answer 1

Draw a diagram to help, it looks something like:

P

/ \

/ \

A __ Q___B

You have two right triangles, which share one leg (the height h = 12 ft).

In related rates problems, you usually need an equation which relates the two quantities. Here you'll have two equations, coming from the Pythagorean theorem.

Let the distance from A to Q be denoted x, and the distance from B to Q be denoted y. There is also a rope over pulley P, of total length 39 ft. Let us denote by T the length from A to P, and therefore 39 - T is the distance from B to P.

We have been told that x' = 1, and wish to find y' at the moment when x = 5.

Notice that when x = 5, we have the two legs in that left triangle as lengths 5 and 12, which tells us what T would be at that moment. (T² = 5² + 12², i.e. T = 13 at that moment.)

Using this value of T, we can find the distance y at that particular moment, which we'll need soon. The right side triangle would satisfy (39 - T)² = y² + 12².

Since T = 13 at that moment, we have 26² = y² + 12² leading to a value of y = 23.06 ft.

Now we can begin taking derivatives. The two relevant equations are:

x² + 12² = T²

y² + 12² = (39 - T)²

Differentiating yields:

2xx' = 2TT'

2yy' = 2(39 - T)(-T')

Plugging the parameters in, then, we have

2(5)(1) = 2(13)T'

which tells us that T' = 10/26 ft / sec.

Using that value of T' in the second equation gives us

2(23.06)y' = 2(39 - 13)(-10/26)

Solving for y' now gives -10/23.06 ft /sec