SOLUTION: Patty and Tom live 345 miles apart. they decide to drive to meet one another. Patty leaves at noon, traveling at an average rate of 45 mi/h, and Tom leaves at 3:00 pm traveling at

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Question 935662: Patty and Tom live 345 miles apart. they decide to drive to meet one another. Patty leaves at noon, traveling at an average rate of 45 mi/h, and Tom leaves at 3:00 pm traveling at an average speed of 60 mi/h. At what time will they meet?
Found 2 solutions by josmiceli, TimothyLamb:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
How far does Patty get in the +3+ hours
before Toby leaves?
+d%5B1%5D+=+45%2A3+
+d%5B1%5D+=+135+ mi
-------------------
Start a stopwatch when Toby leaves
Let +t+ = time in hours on
stopwatch when they meet
Let +d+ = distance in miles that
Toby travels until they meet
-------------------------
Patty's equation:
(1) +345+-d+-+135+=+45t+
Toby's equation:
(2) +d+=+60t+
----------------------
Substitute (2) into (1)
(1) +345+-+60t+-+135+=+45t+
(1) +105t+=+210+
(1) +t+=+2+
They will meet 2 hrs after Toby leaves
3 PM + 2 hrs = 5 PM
is when they meet
---------------------
check:
(2) +d+=+60t+
(2) +d+=+60%2A2+
(2) +d+=+120+ mi
and
(1) +345+-d+-+135+=+45t+
(1) +345+-120+-+135+=+45%2A2+
(1) +90+=+90+
OK

Answer by TimothyLamb(4379) About Me  (Show Source):
You can put this solution on YOUR website!
---
s = d/t
t = d/s
d = st
---
a = distance traveled by patty when tom departs:
a = 45*3
a = 135
---
b = distance between them when tom departs:
b = 345 - a
b = 345 - 135
b = 210
---
speed of patty and tom relative to each other:
s = 45 + 60
s = 105
---
time after tom departs when they meet:
t = d/s
t = 210/105
t = 2 hours
---
answer:
time when they meet:
3:00 pm + 2 hours = 5:00 pm
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