How Far From the Fountain Do the Joggers Meet?

[a.k]

Homework Statement

Two joggers are approaching a water fountain from opposite directions. Jogger A begins 3.0 miles west of the water fountain, with a constant velocity of 4.0 mph due east. Jogger B begins 2.0 miles east of the water fountain, with a constant velocity of 3.0 mph due west. How far are the joggers from the water fountain when they meet one another?

Homework Equations

t=d/v

The Attempt at a Solution

x_a=3 miles west @ 4mph
x_b=2 miles @ 3 mph

t=5/7

I need some guidance as to what I should do now.

 

[tiny-tim]

hi a.k!

t=d/v

The Attempt at a Solution

x_a=3 miles west @ 4mph
x_b=2 miles @ 3 mph

t=5/7

fine so far! …

you've found that, with a relative speed of 7 mph, and a distance of 5 miles, the time taken to reduce the distance to 0 is 5/7 hour

sooo … starting at either end (it makes no difference), at what point will they meet? ​

 

[a.k]

I still don't understand.

 

[tiny-tim]

do you mean you didn't understand this bit? …

you've found that, with a relative speed of 7 mph, and a distance of 5 miles, the time taken to reduce the distance to 0 is 5/7 hour

 

[a.k]

Unfortunately so.

I am curious if this is correct:

5/7*4mph and 5/7*3mph

20/7 and 15/7

2.86 and 2.14 miles

2.86-2.14= 0.14 miles west of the fountain

I think I got this by a fluke by using d=tv then subtracted the mileage.

 

[tiny-tim]

5/7*4mph and 5/7*3mph

20/7 and 15/7

2.86 and 2.14 miles

2.86-2.14= 0.14 miles west of the fountain.

yes this is correct

21/7 miles minus 20/7 miles = 1/7 miles west

14/7 miles minus 15/7 miles = minus 1/7 miles east = 1/7 miles west

either way you get 1/7 miles west

i don't understand what you're not getting about this ​

 

[a.k]

I don't know either tiny tim.

I am happy I was able to figure it out though.

 

[majormaaz]

you've found that, with a relative speed of 7 mph, and a distance of 5 miles, the time taken to reduce the distance to 0 is 5/7 hour

I still don't understand.

Basically, your total distance is 2 + 3 miles = 5 miles, and speed is 4 - (-3) = 7 mph. You have distance, speed and time. Surely you know the equation that makes it all work.

Anyways, your most recent work looks fine.