Question

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 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 (in ft/s) is cart b moving toward q at the instant when cart a is 5 ft from q? (round your answer to two decimal places.) ft/s resources read it

Explanation:

Step1: Establish the relationship using Pythagorean theorem

Let the distance of cart A from Q be $x$, the distance of cart B from Q be $y$, and the length of the rope is a constant $L = 39$ ft. The height from the pulley to the floor is $h = 12$ ft. By the Pythagorean theorem, $\sqrt{x^{2}+12^{2}}+\sqrt{y^{2}+12^{2}}=39$.
When $x = 5$, we first find the length of the rope from cart A to the pulley. Using the Pythagorean theorem, $l_{A}=\sqrt{5^{2}+12^{2}}=\sqrt{25 + 144}=\sqrt{169}=13$ ft. Then the length of the rope from cart B to the pulley is $l_{B}=39 - 13=26$ ft. And we can find $y$ using the Pythagorean theorem for cart B: $y=\sqrt{26^{2}-12^{2}}=\sqrt{(26 + 12)(26 - 12)}=\sqrt{38\times14}=\sqrt{532}\approx23.06$ ft.

Step2: Differentiate the Pythagorean - related equation with respect to time $t$

Differentiating $\sqrt{x^{2}+144}+\sqrt{y^{2}+144}=39$ with respect to $t$ gives:
$\frac{x}{\sqrt{x^{2}+144}}\frac{dx}{dt}+\frac{y}{\sqrt{y^{2}+144}}\frac{dy}{dt}=0$.
We know that $\frac{dx}{dt}=2$ ft/s, $x = 5$, and $y\approx23.06$.
Substitute these values into the differentiated equation:
$\frac{5}{\sqrt{5^{2}+144}}\times2+\frac{23.06}{\sqrt{23.06^{2}+144}}\frac{dy}{dt}=0$.
Since $\frac{5}{\sqrt{169}}\times2+\frac{23.06}{\sqrt{532}}\frac{dy}{dt}=0$, and $\frac{5}{13}\times2+\frac{23.06}{\sqrt{532}}\frac{dy}{dt}=0$, $\frac{10}{13}+\frac{23.06}{\sqrt{532}}\frac{dy}{dt}=0$.
Solve for $\frac{dy}{dt}$:
$\frac{23.06}{\sqrt{532}}\frac{dy}{dt}=-\frac{10}{13}$.
$\frac{dy}{dt}=-\frac{10}{13}\times\frac{\sqrt{532}}{23.06}$.
$\frac{dy}{dt}\approx - 0.61$ ft/s. The negative sign indicates that cart B is moving toward Q.

Answer:

$0.61$