'Monk and the Monastery' problem

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AI Thread Summary
The discussion revolves around determining if there is a point along the monk's path where he passes the same position at the same time on two different days. Participants explore equations to represent the monk's position over time, specifically focusing on the relationship between distance, velocity, and time. They emphasize the need to clarify variables such as initial position (x0), final position (xf), and time (t) in their calculations. The conversation highlights the importance of consistent velocity and how to express it mathematically. Ultimately, the group is working towards simplifying their formulas to find a solution to the problem.
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Homework Statement


Check attached image. Is there any point along the path that he passes at the same time each day

Homework Equations


None that I know of...

The Attempt at a Solution


The whiteboard image, I'm not sure if that's how the graph is suppose to look like.
 

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I do believe that there is a point where the monk would be at the same position at the same time at on both days, but how would I explain it without" saying look at my graph"?
 
Kingyou123 said:
I do believe that there is a point where the monk would be at the same position at the same time at on both days, but how would I explain it without" saying look at my graph"?
Let's try to write down some equations. Suppose that you mark off time as hours since noon and you mark off position as current distance from the city. Can you write down an equation for the man's position in terms of time during the trip from city to monastery?
 
xf=x0+v0(t) t=hours and x= position
 
Kingyou123 said:
xf=x0+v0(t) t=hours and x= position
Can you write down v0 in terms of D?
 
If the velocity constant then the [v][/0] would be D/t
 
Kingyou123 said:
If the velocity constant then the [v][/0] would be D/t
You may want to rethink that. In the equation you gave, t is a variable. But velocity is a constant.
 
xf=xo+x(t)?
 
Kingyou123 said:
xf=xo+x(t)?
Hint: You know how long it takes to get from the city to the monastery.
 
  • #10
jbriggs444 said:
Hint: You know how long it takes to get from the city to the monastery.
xf=xo+x(7), 7 is the number of hours from the office to the monastery.
 
Last edited:
  • #11
Kingyou123 said:
xf=xo+x(7), 7 is the number of hours from the office to the monastery.
The problem statement specifies the value for xf.
 
  • #12
jbriggs444 said:
The problem statement specifies the value for xf.
D=0+v(7) would the second equation be D=D+v(3)?
 
  • #13
Kingyou123 said:
D=0+v(7) would the second equation be D=D+v(3)?
The thing that confuses me is what to solve for, D?
 
  • #14
Refer to post #5.
 
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  • #15
jbriggs444 said:
Refer to post #5.
V0=(xf-x0)/t
 
  • #16
Kingyou123 said:
V0=(xf-x0)/t
What is x0? What is xf? And, for that matter, what is t?
 
  • #17
jbriggs444 said:
What is x0? What is xf? And, for that matter, what is t?
t= times from office to monastery or time from monastery to office
xf=is the distance between the monastery and office
x0=is zero since its the starting distance
 
  • #18
Kingyou123 said:
t= times from office to monastery or time from monastery to office
Given that t=7 and xf-x0 = D, can you simplify the formula for the velocity on the trip from office to monastery?
 
  • #19
I already figure out that step the two formulas set equal to each other would be (D/7)(t)=D-(D/3)t. My answer is 2.1 hours, I'm just having trouble finding the formula to plug it in. Sorry for being confused, in post #6 I had v0= D/t and I thought you meant it was wrong so I completely threw out that answer.
 
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