Racing Runners
Two runners compete in a race of considerable distance. The first runs for a period which is half his total time, then walks the remainder . The second runs half the distance, then walks the second half.
Both begin at the same time ... run the same speed ... and walk the same speed.
Who wins the race?
What if both racers walk first? With all other conditions the same, who wins?
Both begin at the same time ... run the same speed ... and walk the same speed.
Who wins the race?
What if both racers walk first? With all other conditions the same, who wins?
posted by Zaux at
10:19 AM
8 Comments:
The first runner will win.
This is because the second runner will run for less than half his total time, as he will cover the first half of the course in less time than the second.
As the First runner spends longer at a quicker pace he will have a faster average and therefore a shorter time.
It makes no difference whether the racers run or walk first.
This post has been removed by the author.
Repost as made some major sillies.
Part 1. Let d be the total distance. r the running speed, w the walking speed.
For racer 1, d = r*t1/2 + w*t1/2, where t1 is the total time.
=> t1 = 2d/(r+w)
For racer 2, d/2 = w*t21 = r*t22
So, t2 = t21 + t22 = d/(2w) + d/(2r) = d(r+w)/(2rw)
So t2-t1 = d[(r+w)/(2rw) - 2/(r+w)]
= d[(r+w)² - 4rw]/(2rw(r+w))
= d[r²+ 2rw + w² -4rw)]/(2rw(r+w))
= d(r-w)²/(2rw(r+w)) > 0
So t2 > t1 and so runner 1 wins the race.
Part 2: Except that the runners swap run with walk, there is no dif
So t2 > t1 and so runner 1 wins the race.
Thanks Zaux, I was able to comprhend the English this time ;)
Racing Runners
d=v/t
if t of both legs is equal
d=(v1+v2)*t
and t=d/(v1+v2)
call this tA
if d of both legs is equal
t1=d/(2*v1), t2=d/(2*v2)
tB=t1+t2=d*(1/(2*v1)+1/(2*v2))
check the ratio of times
tA/tB=[d/(v1+v2)]/[ d*(1/(2*v1)+1/(2*v2))]
simplify 1/((2*v1)+1/(2*v2))=(2*v1+2*v2)/(4*v1*v2)
tA/tB=4*v1*v2/(2*(v1+v2)^2)
tA/tB=(2*v1*v2)/(v1+v2)^2
let v2 be a ratio R of v1, R>1 (v2 running speed is > v1 walking)
tA/tB=2*R*v1^2/((1+R)*v1)^2
tA/tB=2*R/(1+R)^2
(R+1)^2=R^2+2R+1, so (R+1)^2-2*R=R^2+1, which is >0 for all real R, thus bottom is larger than top
tA/tB <1 for all R>1
so tA <tB and the first runner wins
The equations are not dependent on whether the walking portion occurs first or second.
Answer:
-First runner wins
-The answer remains true if they both walk first. i.e First runner wins
Cam
Cam, I can't let you get away with that. You definitely need more coffee ;)
tA = 2*t => you should have got tA/tB = 4*v1*v2/(v1+v2)^2. Now, your ratio argument needs further work.
(1+R)^2 - (1-R)^2 = 4R
Then get ((1+R)^2 - (1-R)^2)/(1+R)^2
= 1 - ((1-R)/(1+R))^2 < 1 (for R != 1) as required.
So it doesn't matter if they walk faster than they run (if I haven't goofed). Too avoid negative times, require R >= 0.
I hadn't seen, earlier, how to get the required ratio argument, that's why I went for the time difference.
Please don't be upset with me - at least you know someone it is thoroughly appreciating your work.
tA/tB plots (well worth a look):
http://www.wolframalpha.com/input/?i=Plot%5B1-%28%281-R%29%2F%281%2BR%29%29%5E2%5D
Reposting after correcting error identified by Chris.
Racing Runners
d=v/t
if t of both legs is equal
d=(v1+v2)*t*1/2
and t=2*d/(v1+v2)
call this tA
if d of both legs is equal
t1=d/(2*v1), t2=d/(2*v2)
tB=t1+t2=d*(1/(2*v1)+1/(2*v2))
check the ratio of times
tA/tB=[2*d/(v1+v2)]/[ d*(1/(2*v1)+1/(2*v2))]
simplify 1/((2*v1)+1/(2*v2))=(2*v1+2*v2)/(4*v1*v2)
tA/tB=2*4*v1*v2/(2*(v1+v2)^2)
tA/tB=(4*v1*v2)/(v1+v2)^2
let v2 be a ratio R of v1, R>1 (v2 running speed is > v1 walking)
tA/tB=4*R*v1^2/((1+R)*v1)^2
tA/tB=4*R/(1+R)^2
(R+1)^2=R^2+2R+1, so (R+1)^2-4*R=R^2-2*R+1,or (R-1)^2 which is >0 for R>1, thus bottom is larger than top
tA/tB <1 for all R>1
so tA <tB and the first runner wins
The equations are not dependent on whether the walking portion occurs first or second.
Answer:
-First runner wins
-The answer remains true if they both walk first. i.e First runner wins
Cam
Hi Cam. You trumped me. i.e. your ratio argument was better than mine.
Post a Comment
Links to this post:
Create a Link
<< Home