SOLUTION: brittany can run 1 mile in 6 min and christina can run 1mile in 9 min. if both brittany and christiana start at the same time nd brittany starts 1/3 of a mile behind christina how

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Question 227176: brittany can run 1 mile in 6 min and christina can run 1mile in 9 min. if both brittany and christiana start at the same time nd brittany starts 1/3 of a mile behind christina how long will it take brittany to catch up to christina
Found 2 solutions by ankor@dixie-net.com, josmiceli:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Brittany can run 1 mile in 6 min, and Christina can run 1 mile in 9 min.
If both Brittany and Christina start at the same time, and brittany starts 1/3 of
a mile behind Christina, how long will it take Brittany to catch up to Christina
:
I may be easier to do this in ft/min
B's speed: 5280%2F6 = 880 ft/min
C's speed: 5280%2F9 = 586.67 ft/min
:
1%2F3*5280 = 1760 ft is 1/3 of a mile
:
Let t = time (in minutes) for B to catch C
:
Write a distance equation (dist = speed * time)
;
B's dist = C's dist + 1/3 mile
880t = 586.67t + 1760
:
880t - 586.67t = 1760
:
293.33t = 1760
t = 1760%2F293.33
t = 6 minutes
:
:
Check solution by finding the dist of each
6*880 = 5280
6*586.67 = 3520
---------------
difference 1760, which is 1/3 of a mile

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
They will both be running for the same
amount of time. They just have different
starting points.
Let t = the amount of time they will
be running in min
Each one has an equation.
Christina:
d%5Bc%5D+=+r%5Bc%5D%2At
Brittany:
d%5Bb%5D+=+r%5Bb%5D%2At
given:
r%5Bc%5D+=+1%2F9 mi/min
r%5Bb%5D+=+1%2F6 mi/min
d%5Bb%5D+=+d%5Bc%5D+%2B+1%2F3
-----------------------
rewriting equations:
(1) d%5Bc%5D+=+%281%2F9%29%2At
(2) d%5Bc%5D+%2B+1%2F3+=+%281%2F6%29%2At
I have 2 equations and 2 unknowns, so it's solvable
(1) d%5Bc%5D+=+%281%2F9%29%2At
9d%5Bc%5D+=+t
and
(2) d%5Bc%5D+%2B+1%2F3+=+%281%2F6%29%2At
6d%5Bc%5D+%2B+2+=+t
Substitute (1) in (2)
6d%5Bc%5D+%2B+2+=+9d%5Bc%5D
3d%5Bc%5D+=+2
d%5Bc%5D+=+2%2F3 mi
Plugging back into (1)
(1) d%5Bc%5D+=+%281%2F9%29%2At
2%2F3+=+%281%2F9%29%2At
t+=+9%2A%282%2F3%29
t+=+6 min
It will take Brittany 6 min to catch Christina
check:
(1) d%5Bc%5D+=+%281%2F9%29%2At
2%2F3+=+%281%2F9%29%2A6
2%2F3+=+6%2F9
2%2F3+=+2%2F3
and
(2) d%5Bc%5D+%2B+1%2F3+=+%281%2F6%29%2At
2%2F3+%2B+1%2F3+=+%281%2F6%29%2A6
1+=+1
OK