Question

A railway system on a hillside moves passengers at a constant rate to an elevation of 50 m. The elevations of a train are given for 2 different locations.

a. Write an equation in​ point-slope form to represent the elevation of the train in terms of time. How can the equation be used to find the rate of increase in elevation of the train in meters per​ second?
b. At what elevation does the train start​ initially? Write a linear equation in a form that gives the information as part of the equation. Explain your reasoning.

Answer 1

To determine the train's elevation over time at a constant rate of increase, use the point-slope form of a linear equation and rearrange the slope to find the rate. If the train starts at an initial elevation of zero, the elevation equation is simply e = rt.

Step-by-step explanation:

To write an equation in point-slope form for the elevation of the train given it moves at a constant rate to an elevation of 50 meters, you'll need two things: a specific time-elevation point and the rate of increase in elevation per unit of time. Let's denote the time by t (in seconds, for instance), and the elevation by e (in meters). If we know the train's elevation at a certain time—say, e1 at time t1—and we assume it starts at ground level (0 meters) at t=0, then the slope (rate of elevation) is the change in elevation over the change in time (Δe/Δt=50/t). If 50 meters is reached at t1, the point-slope form is e - e1 = (e1/t1)(t - t1). This equation helps to find the rate of increase by rearranging it to Δe/Δt = slope.

For part b, if we know the train begins its journey at ground level, the linear equation that gives the elevation at any time t would simply be e = (r)(t), where r is the constant rate of elevation. If, say, the train reaches 50m in 200 seconds, then r would be 50/200 = 0.25 meters per second and the equation would be e = 0.25t.