Question

A conveyor belt carries supplies from the first floor to the second floor, which is 12 feet higher. The belt makes a 60° angle with the ground. How far do the supplies travel from one end of the conveyor belt to the other? If the belt moves at 80 ft / min how long does it take the supplies to move to the second floor to the nearest tenth of a minute.
A. 17 ft ; 0.2 min
B. 35 ft ; 0.4 min
C. 42 ft ; 0.5 min
D. 52 ft ; 0.6 min

Answer 1

We can use trigonometry to solve this problem. Let's call the distance the supplies travel "x". Then we can set up a right triangle with the conveyor belt as the hypotenuse, and the ground and the vertical height of the second floor as the legs:

[asy]
pair A=(0,0), B=(43.3,25), C=(43.3,0);
draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,60));
label("60°",C+(1,0));
label("$x$ ft",B/2);
label("12 ft",B-C/2);
label("$h$ ft",C/2);
[/asy]

Using the angle and the vertical leg, we can find the length of the hypotenuse:

$$\sin 60^\circ = \frac{h}{x} \quad \Rightarrow \quad x = \frac{h}{\sin 60^\circ} = \frac{12}{\sqrt{3}/2} = 24\sqrt{3} \approx 41.6 \text{ ft}$$

So the supplies travel about 41.6 feet from one end of the conveyor belt to the other. To find how long it takes for the supplies to move to the second floor, we can divide the distance by the speed of the belt:

$$\frac{41.6 \text{ ft}}{80 \text{ ft/min}} \approx 0.52 \text{ min} \approx 0.5\text{ min}$$

So the answer is (C) 42 ft ; 0.5 min.