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
a. If x is the time in seconds and y is the elevation of the train in meters, then represents the elevation of the train in terms of time.
(Type an equation. Type your answer in point-slope form. Use integers or decimals for any numbers in the equation.)

Answer 1

Answer:

part a) y

-9=0.4(x-14.5)

part b) slope= 0.4

part c) 3.2 meters

final answer is y=0.4x+3.2( slope-intercept )

Step-by-step explanation:

hope that help :)

Answer 2

The equation of the model given written in point slope form is :

• y - 9 = 0.4(x - 14.5)

• Velocity on meter per second = 0.4 m/s

• Initial elevation = 3.2 m

Final elevation = 50 m

Using the information given :

9 meter = 14.5 seconds - - - - (1)

20 meter = 42 seconds - - - - (2)

(20 - 9)meters = (42 - 14.5) seconds

11 meters = 27.5 seconds

• Speed = distance / time

Speed = 11 / 27.5

Speed = 0.4 meters per second

The equation in point slope form can be represented thus :

The initial velocity :

At y = 9 meters ; x = 14.5 seconds ; c = initial elevation

y = 0.4x + c

9 = 0.4(14.5) + c

9 = 5.8 + c

c = 9 - 5.8

c = 3.2 m

In point - slope form, the equation can be written as :

Therefore, the equation of the model written in point - slope form is : y - 9 = 0.4(x - 14.5)

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