What time did they both set off?

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Homework Help Overview

The problem involves two individuals, Willy and Milly, who leave from different points and travel towards each other at constant speeds. They meet at noon, and the task is to determine the time they both set off based on their arrival times at each other's starting points.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationships between distance, speed, and time, and how to set up equations based on the given information. There is an exploration of substituting variables in equations to find unknowns.

Discussion Status

The discussion is ongoing, with participants examining the equations derived from the problem. Some guidance has been provided regarding the substitution of variables to simplify the equations, and there is acknowledgment of the assumption that both individuals set off at the same time.

Contextual Notes

There is a recognition that the problem involves three unknowns, which may complicate finding a numerical solution. Participants are navigating the implications of their assumptions and the information provided in the problem statement.

GoghAway

Homework Statement


Willy leaves point A and Milly leaves point B, both of them are heading toward each other's starting point. They both have constant speeds, but Milly is faster, and they pass each other at noon without stopping. Milly arrives at point A at 1:00 pm, and Willy arrives at point B at 2:15 pm. At what time did they both set off?

Homework Equations


v = d / t
v = (d2 - d1) / (t2 - t1)

The Attempt at a Solution


I'm putting the data here because it's what I think the problem is telling me, but I know I could be wrong.

Willy
d = D
v = Vw
t = (14.25 hours - t1)

Vw = D / (14.25 h - t1)

Milly
d = D
v = Vm
t = (13.0 hours - t1)

Vm = D / (13.0 h - t1)

After noon
Willy
d = (D - d1)
v = Vw
t = 2.25 hours

Vw = (D - d1) / 2.25 h

Milly
d = d1
v = Vm
t = 1 hour

Vm = d1 / (1 h)

Since Milly is going faster noon, or when they meet would be after the midpoint between point A and B for her, and before the midpoint for Willy.

Since Willy's velocity is always the same:
D/(14.25 h - t1) = (D - d1)/2.25 h
D (2.25 h) = (D - d1)(14.25 h - t1)
D = ((D - d1)(14.25 h - t1)) / (2.25 h)

Since Milly's velocity is always the same:
D/(13.0 h - t1) = d1/(1 h)
D (1 h) = d1 (13.0 h - t1)
D = (d1 (13.0 h - t1)) / (1 h)
d1= (D (1 h)) / (13.0 h - t1)


I'm not sure whether I can sub what D = or what d1 = from what I have since Willy's velocity is always the same with what I have since Milly's velocity is always the same or vice versa. Either way, I don't think I know enough of the unknowns to come up with a numerical value as an answer, but I don't know how to get the information that I need.
 
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Welcome to PF.

Your method will work. You have gotten to the two equations

D = ((D - d1)(14.25 h - t1)) / (2.25 h)

d1= (D (1 h)) / (13.0 h - t1)

It looks like you have two equations with three unknowns. But see what happens if you substitute the expression for d1 of the second equation into the first equation.
 
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I believe we are given that they both set off at the same time.
 
Thank you TSny! I did as you suggested, and checked the times to see if both Milly and Willy had the correct amount of time after they meet at noon so they would arrive at the correct times, which they did!
 

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