Modeling simple linear equations. This should be pretty simple but I'm blanking on this. I need to model (graph) how path 1 becomes equally as efficient as path 2 as the distance of path 2 increases.

Josie Kennedy

Josie Kennedy

Answered question

2022-11-02

Modeling simple linear equations
This should be pretty simple but I'm blanking on this. I need to model (graph) how path 1 becomes equally as efficient as path 2 as the distance of path 2 increases.
distance of path 1 (from A to B) = 10
distance of path 2 (from A to C) = y
The path is a path taken by a forklift driver. The vehicle speed is 2.2m/s. The paths are measured in meters.
On path 1, there is an added flat time of 30 seconds (due to something he has to stop and do). On path 2, there is an added flat time of 10 seconds.
So in summary:
x 2.2 + 30 = t path  1           w h e r e       x > 0
y 2.2 + 10 = t path  2           w h e r e       y > 0
I know that once the distance of path 2 = 54 meters, both routes take the same amount of time (34.545 seconds). But I can't figure out how to put this into equations that I can use to graph.

Answer & Explanation

Maffei2el

Maffei2el

Beginner2022-11-03Added 20 answers

Step 1
Color scheme: time, distance, velocity
Given a velocity v, the time to go from A to B is
t 1 ( v ) = 10 v + 30 .
The time to go from A to C is
t 2 ( v ) = y v + 10
When y is small t 2 < t 1 ; when y is larger, t 2 > t 1 .
Step 2
Find the value of y for which
(1) t 1 ( v ) = t 2 ( v )
Solving (1),
t 1 ( v ) = t 2 ( v ) 10 v + 30 = y v + 10
The distance is
y = 10 + 20 v
Given v = 2.2, the distance is y = 54 .

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?