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. 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.
Answers
GPT-4o mini
To develop the equations and insights regarding the railway system on a hillside, we need to establish a few parameters:
### Given Information:
- The elevation of the train, \( y \), is ultimately 50 m.
- The rate of increase in elevation is constant.
### Part (a): Point-Slope Form Equation
To write an equation in point-slope form, we need a reference point and the rate of elevation increase (slope).
#### 1. Determine Points
Let's assume we have two points that we refer to:
- First point: At time \( t_1 \), suppose the elevation is \( y_1 \).
- Second point: At time \( t_2 \), the elevation is at its final \( y_2 = 50 \) m.
Let’s denote the elevation at time \( t_1 \) as \( y_1 = 0 \) m (this is the initial elevation).
#### 2. Calculate the Slope
The slope (rate of increase in elevation) is given by:
\[ m = \frac{y_2 - y_1}{t_2 - t_1} \]
Assuming \( y_1 = 0 \) m and \( y_2 = 50 \) m, we can substitute known values:
\[ m = \frac{50 - 0}{t_2 - t_1} = \frac{50}{t_2 - t_1} \]
Let’s denote \( t_2 - t_1 \) as \( t \).
Thus, the slope simplifies to \( m = \frac{50}{t} \).
#### 3. Point-Slope Form Equation
Using point-slope form, the equation can be structured as:
\[
y - y_1 = m(t - t_1)
\]
Substituting the values we have:
\[
y - 0 = \frac{50}{t}(t - t_1)
\]
This simplifies to:
\[
y = \frac{50}{t}(t - t_1)
\]
##### **Using the Equation to Find Rate of Elevation**
The slope \( m \) (i.e., \( \frac{dy}{dt} \)) tells us how much the elevation increases per second. Therefore, it can be directly interpreted as the rate of increase in elevation in meters per second.
### Part (b): Initial Elevation and Linear Equation
The elevation where the train starts initially is \( y_1 \) which we assumed to be \( 0 \) m. Therefore, the train starts its journey from a base elevation of \( 0 \) meters before ascending to \( 50 \) meters.
#### Write a Linear Equation
The general form of a linear equation representing this scenario with a known starting point (initial elevation), the maximum elevation, and an understood relationship can be structured as:
\[
y = mx + b
\]
where:
- \( m = \) slope (rate of increase, which we found as \( \frac{50}{t} \)).
- \( b = 0 \) is the initial elevation at \( t = 0 \).
Thus, the corresponding linear equation will be:
\[
y = \frac{50}{t}t + 0
\]
Since we don’t know \( t \) yet, and because the linear relationship essentially models the rise from \( 0 \) to \( 50 \) m over the time interval.
### Reasoning Explanation
- The point-slope form makes it clear that the elevation starts at zero and increases based on a constant rate calculated from the time it takes to reach the final elevation of 50 m.
- The train's route can be linearized based on consistent increments in time and height.
This verbal and mathematical reasoning ties the physical reality of the train's journey to its representation through basic algebraic forms.
### Given Information:
- The elevation of the train, \( y \), is ultimately 50 m.
- The rate of increase in elevation is constant.
### Part (a): Point-Slope Form Equation
To write an equation in point-slope form, we need a reference point and the rate of elevation increase (slope).
#### 1. Determine Points
Let's assume we have two points that we refer to:
- First point: At time \( t_1 \), suppose the elevation is \( y_1 \).
- Second point: At time \( t_2 \), the elevation is at its final \( y_2 = 50 \) m.
Let’s denote the elevation at time \( t_1 \) as \( y_1 = 0 \) m (this is the initial elevation).
#### 2. Calculate the Slope
The slope (rate of increase in elevation) is given by:
\[ m = \frac{y_2 - y_1}{t_2 - t_1} \]
Assuming \( y_1 = 0 \) m and \( y_2 = 50 \) m, we can substitute known values:
\[ m = \frac{50 - 0}{t_2 - t_1} = \frac{50}{t_2 - t_1} \]
Let’s denote \( t_2 - t_1 \) as \( t \).
Thus, the slope simplifies to \( m = \frac{50}{t} \).
#### 3. Point-Slope Form Equation
Using point-slope form, the equation can be structured as:
\[
y - y_1 = m(t - t_1)
\]
Substituting the values we have:
\[
y - 0 = \frac{50}{t}(t - t_1)
\]
This simplifies to:
\[
y = \frac{50}{t}(t - t_1)
\]
##### **Using the Equation to Find Rate of Elevation**
The slope \( m \) (i.e., \( \frac{dy}{dt} \)) tells us how much the elevation increases per second. Therefore, it can be directly interpreted as the rate of increase in elevation in meters per second.
### Part (b): Initial Elevation and Linear Equation
The elevation where the train starts initially is \( y_1 \) which we assumed to be \( 0 \) m. Therefore, the train starts its journey from a base elevation of \( 0 \) meters before ascending to \( 50 \) meters.
#### Write a Linear Equation
The general form of a linear equation representing this scenario with a known starting point (initial elevation), the maximum elevation, and an understood relationship can be structured as:
\[
y = mx + b
\]
where:
- \( m = \) slope (rate of increase, which we found as \( \frac{50}{t} \)).
- \( b = 0 \) is the initial elevation at \( t = 0 \).
Thus, the corresponding linear equation will be:
\[
y = \frac{50}{t}t + 0
\]
Since we don’t know \( t \) yet, and because the linear relationship essentially models the rise from \( 0 \) to \( 50 \) m over the time interval.
### Reasoning Explanation
- The point-slope form makes it clear that the elevation starts at zero and increases based on a constant rate calculated from the time it takes to reach the final elevation of 50 m.
- The train's route can be linearized based on consistent increments in time and height.
This verbal and mathematical reasoning ties the physical reality of the train's journey to its representation through basic algebraic forms.