SOLUTION: A ship left port A with a speed of 40 mph. In 1 hour and 15 minutes, a second ship left the port with a speed of 60 mph. In how many hours and at what distance from port A will the

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A ship left port A with a speed of 40 mph. In 1 hour and 15 minutes, a second ship left the port with a speed of 60 mph. In how many hours and at what distance from port A will the      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 955829: A ship left port A with a speed of 40 mph. In 1 hour and 15 minutes, a second ship left the port with a speed of 60 mph. In how many hours and at what distance from port A will the second ship catch up with the first one?
Found 2 solutions by josmiceli, stanbon:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
In +1.25+ hrs, the 1st ship travels:
+d%5B1%5D+=+40%2A1.25+
+d%5B1%5D+=+50+ mi
-----------------
Start a stopwatch when 2nd ship leaves
Let +t+ = time in hrs when 2nd ship
catches the 1st ship
Let +d+ = distance from port A when
2nd ship catches 1st ship
-----------------------
Equation for 1st ship:
(1) +d+-+50+=+40t+
Equation for 2nd ship:
(2) +d+=+60t+
----------------------
Substitute (2) into (1)
(1) +60t+-+50+=+40t+
(1) +20t+=+50+
(1) +t+=+2.5+
------------------
and
(2) +d+=+60%2A2.5+
(2) +d+=+150+
150 mi from port A, 2nd ship catches the 1st ship
---------------
check:
(1) +d+-+50+=+40t+
(1) +150+-+50+=+40%2A2.5+
(1) +100+=+100+
OK




Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
A ship left port A with a speed of 40 mph. In 1 hour and 15 minutes, a second ship left the port with a speed of 60 mph. In how many hours and at what distance from port A will the second ship catch up with the first one?
------------------
1st ship DATA:
rate = 40 mph ; time = x hrs ; distance = 40x miles
----
2nd ship DATA:
rate = 60 mph ; time = (x-(5/4)) hrs ; distance = 60(x-(5/4)) miles
---------
Equation::
dist = dist
40x = 60x - 75
20x = 75
x = 15/4 hrs
----
Dist from 1st port:: 40x = 40(15/4) = 150 miles
--------------
Cheers,
Stan H.