SOLUTION: Sara leaves at 7am at a rate of 45mph
Mark leaves at 7:30am at a rate of 55mph
They both leave from the same location traveling in the same direction
At what time will Mark catc
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-> SOLUTION: Sara leaves at 7am at a rate of 45mph
Mark leaves at 7:30am at a rate of 55mph
They both leave from the same location traveling in the same direction
At what time will Mark catc
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Question 551725: Sara leaves at 7am at a rate of 45mph
Mark leaves at 7:30am at a rate of 55mph
They both leave from the same location traveling in the same direction
At what time will Mark catch up with Sara?
What equation or equations should be used when solving this problem algebraically? Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! Sara leaves at 7am at a rate of 45mph
Mark leaves at 7:30am at a rate of 55mph
They both leave from the same location traveling in the same direction
At what time will Mark catch up with Sara?
What equation or equations should be used when solving this problem algebraically?
:
Let t = Sara's travel time
then
(t-.5) = Mark's travel time (leaves half hour later)
:
A simple equation can be derived from the fact, when M catches S, they will have traveled the same distance,
Dist = speed * time
:
M's dist = S's dist
55(t-.5) = 45t
55t - 27.5 = 45t
55t - 45t = 27.5
10t = 27.5
t =
t = 2.75 hrs from S's starting time (2 hrs + .75(60) = 45 min)
:
7:00 + 2:45 = 9:45 when M catches S
:
:
We can confirm this by finding the dist each traveled, if we did this right, they should be equal
45*2.75 = 123.75 mi
55*2.25 = 123.75
