   Chapter 3.9, Problem 42E

Chapter
Section
Textbook Problem

Two carts, A and B, are connected by a rope 39 ft long that passes over a pulley P (see the figure). The point Q is on the floor 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 2 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q? To determine

To find: The rate at which the cart B is moving toward the point Q when cart A is 5 ft from Q.

Explanation

Given:

The two carts connected by a rope 39 ft long and pass over the pulley P .

The point Q is on the floor 12 ft directly below the point P and between the carts.

The rate of moving of the cart A away from the point Q is 2 ft/s.

Formula used:

(1) Chain rule:

(2) Pythagorean Theorem.

Calculation:

The two carts A and B connected by the 39 ft long rope and pass over the pulley P.

Let us assume that x be the distance between the cart A and the point Q and y be the distance between the cart B and the point Q as shown the figure 1 given below.

Let consider Q as the origin point.

Since the distance x and y changes with the time t.

Therefore, x and y are the function of the time t.

Since dxdt=2ft/s.

Obtain dydt when x=5ft from Q.

From the above figure 1 the total length is.

AB=AP+PB .

Then by using the Pythagorean Theorem in the triangle AQP and BQP

AB=x2+122+y2+122 .

Since the total length of the rope AB= 39 ft, therefore

x2+122+y2+122=39

Differentiate the total length of the rope with respect to the time t.

ddt[x2+122+y2+122]=ddtddx[x2+122]dxdt+ddy[y2+122]dydt=0[Qdydx=dydududx][12(x2+122)121(2x)]dxdt+[12(y2+122)121(2y)]dydt=0[xx2+122]dxdt+[yy2+122]dydt=0

On further simplification the rate of change speed of the cart Q is as follows

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