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 h = 12 ft directly beneath P and between the carts. Cart A is being pulled away from Q at a speed of 3.5 ft/s. How fast is cart B moving toward Q at the instant when cart A is 5 ft from Q? (Round your answer to two decimal places.)
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To find the rate at which cart B is moving toward Q, we can use related rates.
Let's first assign variables to the information given in the problem:
- Let x represent the distance between cart A and Q.
- Let y represent the distance between cart B and Q.
We are given that cart A is being pulled away from Q at a speed of 3.5 ft/s. This means that dx/dt = 3.5 ft/s. We want to find dy/dt, the rate at which cart B is moving toward Q.
We also know that the two carts are connected by a rope 39 ft long. Therefore, we can write the relationship between x and y using the Pythagorean theorem:
x^2 + y^2 = 39^2
Differentiating both sides of the equation with respect to time (t), we get:
2x(dx/dt) + 2y(dy/dt) = 0
Since we are interested in finding dy/dt when x = 5 ft, we can substitute these values into the equation:
2(5)(3.5) + 2y(dy/dt) = 0
Simplifying the equation gives:
35 + 2y(dy/dt) = 0
Now, we can solve for dy/dt:
2y(dy/dt) = -35
dy/dt = -35 / (2y)
To find the value of y when x = 5 ft, we can use the Pythagorean theorem:
5^2 + y^2 = 39^2
25 + y^2 = 1521
y^2 = 1496
y = sqrt(1496) ≈ 38.652
Substituting this value into the equation for dy/dt, we get:
dy/dt = -35 / (2 * 38.652)
dy/dt ≈ -0.452 ft/s
Therefore, at the instant when cart A is 5 ft from Q, cart B is moving toward Q at a rate of approximately -0.452 ft/s. The negative sign indicates that cart B is moving in the opposite direction of cart A.