Question

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

Answer 1

Answer:$\frac{20}{23} ft/s$

Step-by-step explanation:

Given

Velocity of point A is 2 ft/s

Length of rope is 39 ft

let x be the distance of A from Q and y be the distance of point B from Q

thus From Diagram

$x^2+12^2=L^2$

Differentiating

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+0=2L\cdot \frac{\mathrm{d} L}{\mathrm{d} t}$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=L\cdot \frac{\mathrm{d} L}{\mathrm{d} t}$

at x=5 ft L=13 ft

thus $\frac{\mathrm{d} L}{\mathrm{d} t}=\frac{5\times 2}{13}=\frac{10}{13}$

For point B

$y^2+12^2=(39-L)^2$

Differentiate w.r.t time

$2y\frac{\mathrm{d} y}{\mathrm{d} t}+0=2L\cdot \frac{\mathrm{d} 39-L}{\mathrm{d} t}$

$y\frac{\mathrm{d} y}{\mathrm{d} t}=-(39-L)\cdot \frac{\mathrm{d} (39-L)}{\mathrm{d} t}$

at $x=5\ ft$

$y=23 ft$

thus $\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{-20}{23}ft/s$