# Runner's motion in one dimension

1. Dec 20, 2008

### Elbobo

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
vf = vo + at
avg velocity = (vf + vo) /2
d = vo)t + (1/2) at2

3. 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.

2. Dec 20, 2008

### NBAJam100

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

3. Dec 20, 2008

### MM156

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.

4. Dec 20, 2008

### LowlyPion

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.

5. Dec 20, 2008

### Elbobo

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).

6. Dec 20, 2008

### LowlyPion

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?

7. Jan 7, 2009

### marindo

actually i dont 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 thats my take on it anyways

8. Jan 7, 2009

### LowlyPion

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.

9. Jan 8, 2009

### marindo

yes = ="

10. Sep 19, 2009

### vv_ramirez259

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?