Runner's motion in one dimension

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Homework Help Overview

The problem involves a runner attempting to complete a 10,000-meter run in under 30 minutes, facing a situation where, after 27 minutes, there are still 1,100 meters remaining. The runner needs to determine the time required to accelerate at a specified rate to meet the goal.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the importance of the runner's initial velocity at the 27-minute mark and how it affects the calculations for the remaining distance. There are suggestions to calculate the average velocity over the first part of the run to use as a starting point for the final segment. Some participants propose breaking down the remaining distance into phases of acceleration and constant velocity.

Discussion Status

There is an ongoing exploration of different interpretations regarding the use of average versus instantaneous velocity in the equations. Some participants express confusion about the appropriateness of using average velocity in certain contexts, while others suggest that it is a reasonable assumption given the problem constraints. The discussion remains open with various perspectives being considered.

Contextual Notes

Participants note the challenge of determining the runner's initial velocity and the implications of assumptions made about the runner's speed prior to the final acceleration phase. There is also mention of the need for clarity in problem statements regarding the use of average speed.

Elbobo
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Homework Statement


A runner hopes to complete the 10,000-m run in less than 30.0 min. After exactly 27.0 min, there are still 1100 m to go.

The runner must then accelerate at 0.20 m/s2 for how many seconds in order to achieve the desired time?


Homework Equations


vf = vo + at
avg velocity = (vf + vo) /2
d = vo)t + (1/2) at2
vf2 = vo2 + 2ad


The Attempt at a Solution


Most of my solutions just wind up at 0 = 0 or a = a... In other words, I'm not doing anything wrong, but it's that I can't think of any other way to solve it.

d1 = 8900 m
t = 1620 s
d2 = 1100 m
avg velocity = 8900 / 1620 = A

(I took this route so I could find the instantaneous velocity at t = 1620s, which I though would lead up to another part that could solve the problem)

A = (vf+ vo) / 2
2A - vf = vo

a = [vf - (2A - vf)] / t
a = (2vf - 2A) /t

vf2 = (2A - vf)2 + 2[ (2vf -2A) /t] d1
(which results in)
0 = 0

I've been thinking about this problem for a loooong time, but I can't figure out another way to solve it.
 
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It looks like this problem really depends on the Vo of the runner at t=27min. say if at that point he stopped at then started from 0m/s, then accelerated at .2m/s, the time would be higher than if he "started" at the average Vo of the overall 27 minute trip. Or say he started slow then started going faster towards the end of the 27 minute and had a higher than average Vo at 27 to make up for his lower starting Vo...

I think a good first step would be to find his average velocity though the first 27 minutes of the trip if he was on pace for a 30 min 10k. Take that Velocity to be the Vo of your starting equations for the last 3 minutes... you are given the accel and you know Vf=0. So try going from there
 
Look for the final velocity with 1100/3 minutes. You might be able to use that velocity to look further into acceleration and time. I haven't looked into it further enough though.
 
Basically what you have to do is cover the 1100 m.

This will be made up of 2 phases. t1)acceleration. t2) constant velocity to finish won't it?

1100 m = Vf*t2 + Vavg*t1 +1/2(.2)(t22)

t1 + t2 = 180 secs

And Vf = Vavg + a*t1

But I'm not solving it.
 
Thank you, that worked.

But I don't understand why the average velocity was able to be used in either equation. I thought those equations can only be used with instantaneous velocities (in this case, when t = 1620 s).
 
Elbobo said:
Thank you, that worked.

But I don't understand why the average velocity was able to be used in either equation. I thought those equations can only be used with instantaneous velocities (in this case, when t = 1620 s).

True enough. But they gave you no choice. You have to assume that he was running at average speed. It is the best bet that he was after 27 minutes. It would have been better to state that he was running at average speed in the problem. But ... hey. What do you want perfect problems?
 
actually i don't think its V_avg, but like Initial velocity...

thats why ur "v_avg" was used... V_avg = Vf - Vi/2 ... and through ur previous calculation of 8900m/1620s = 5.49 m/s its obvious its not V_avg...

i could be wrong, but that's my take on it anyways
 
marindo said:
actually i don't think its V_avg, but like Initial velocity...

thats why ur "v_avg" was used... V_avg = Vf - Vi/2 ... and through ur previous calculation of 8900m/1620s = 5.49 m/s its obvious its not V_avg...

i could be wrong, but that's my take on it anyways

The point is that you use his average velocity over the first part as the initial velocity for when he begins his "kick" to final speed for the last part.
 
LowlyPion said:
But ... hey. What do you want perfect problems?
yes = ="
 
  • #10
LowlyPion said:
Basically what you have to do is cover the 1100 m.

This will be made up of 2 phases. t1)acceleration. t2) constant velocity to finish won't it?

1100 m = Vf*t2 + Vavg*t1 +1/2(.2)(t22)

t1 + t2 = 180 secs

And Vf = Vavg + a*t1

But I'm not solving it.
I am trying to solve this problem as well, and my issue is that I don't really know what each variable in the equation represents

Can someone help me?
 

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