Kinematics In One Direction Problem

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Homework Help Overview

The discussion revolves around a kinematics problem involving a runner attempting to complete a 10,000m run within a specified time. The runner has a remaining distance of 1100m after running for 27 minutes and must accelerate to meet the time constraint.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial velocity calculation and the need to determine the time of acceleration. There is confusion regarding the final velocity and the relationship between distance, time, and acceleration. Some participants suggest breaking the remaining time into segments to analyze the problem further.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the use of kinematic equations, but there is still uncertainty about how to approach the acceleration time specifically.

Contextual Notes

Participants note the importance of the given acceleration and the need to carefully consider the details of the problem setup, including the total time and distance remaining.

Mazero
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A runner hopes to compleate the 10,000m run in less than 30.0min. After running at a constant speed for exactly 27.0min there are still 1100m to go. The runner must then accelerate at 0.20 m/s^2 for how many seconds in order to achieve the desired time.

Ill briefly summarize what i tryed...

Init. Velocity = (10,000-1100) /(27min * 60s) = 5.5 m/s
Final Velocity = 1100/(180-t) since t is not know, because i don't know yet how long it will accelerate for

i was then going to use VF^2 = VI^2 + 2ad to find the distance it must accelerate for, and use that to find the time, this seems rather wrong, hence my confusion, but i figured id give my idea anyway
 
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Mazero said:
Init. Velocity = (10,000-1100) /(27min * 60s) = 5.5 m/s
Good. This is the speed of the runner for the first 27 minutes.
Final Velocity = 1100/(180-t) since t is not know, because i don't know yet how long it will accelerate for
Don't worry about the final speed, since that's not asked for. You know the distance and time for the last leg of the run, and you need to find the acceleration. The kinematic equation that relates these three variables is:
[tex]x = v_0 t + 1/2 a t^2[/tex]
 
I think you misunderstood the question, i need to know how long he accelerates, in order to finish in 3 mins, if you read the question more closely youll realize its not as simple as i thought, i immediatly did what you suggested, until someone else pointed out the other details
 
Mazero said:
I think you misunderstood the question, i need to know how long he accelerates, in order to finish in 3 mins, if you read the question more closely youll realize its not as simple as i thought, i immediatly did what you suggested, until someone else pointed out the other details
Oops... my bad. I forgot that the acceleration was given.

So break the remaining 3 minutes into two segments [itex]t_1[/itex] and [itex]t_2[/itex]. Assume the final speed is [itex]v_f[/itex]. So:
[tex]v_f = v_0 + a t_1[/tex]
[tex]t_1 + t_2 = 180[/tex]
I'll leave the last for you: total distance traveled during [itex]t_1 + t_2[/itex] = 1100 m.

I hope that helps.
 

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