Describe elevator trips in parallel shafts. Assume that the passenger and freight elevators move at steady speeds and can start and stop instantly. The trips are nonstop unless otherwise indicated. Both elevators start at the same time. A notation like (4, 600) means that 4 minutes into a trip, the elevator is 600 feet above the bottom of the elevator shaft. Rough sketches on a coordinate system may be helpful.

In the Ellis Building,

Passenger: Starts 400 ft up, travels to (2, 40)

Freight: (0.5, 100) to (3, 250)

a. When, if ever, are the two elevators at the same height during this trip? What height is that?

b. How fast does each elevator travel?

Solution

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To determine the height where the two elevators will be the same, write an equation that shows the relationship between the time, it takes for the elevator to reach a certain height.

First, write an equation that represents the elevator for the passenger. Let $x_1=0$ , $x_2=2$ , $y_1=400$ and $y_2=40$ . Substitute these values on the equation of the line passing through two points then find the equation needed.

\begin{aligned} y-y_1&=\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1)\\\\ y-400&=\left(\frac{40-400}{2-0}\right)(x-0)\\\\ y-400&=\left(\frac{-360}{2}\right)x\\\\ y-400&=-180x\\\\ y&=-180x+400 \end{aligned}

Next, write an equation that represents the elevator for the freight. Let $x_1=0.5$ , $x_2=3$ $y_1=100$ and $y_2=250$ . Substitute these values on the equation of the line passing through two points then find the equation needed.