SOLUTION: At 12:00 noon Ship A is 80 miles due north of Ship B and is sailing north at a rate of 13 knots. Ship B is sailing east at a rate of 11 knots. Write the distance between the ship

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Question 174316: At 12:00 noon Ship A is 80 miles due north of Ship B and is sailing north at a rate of 13 knots. Ship B is sailing east at a rate of 11 knots. Write the distance between the ships as a function of time, where T=0 represents 12:00 noon and determine how far apart (in miles) the two ships will be at 5pm
(1 Knot = 1.15mph)
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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At 12:00 noon Ship A is 80 miles due north of Ship B and is sailing north
at a rate of 13 knots. Ship B is sailing east at a rate of 11 knots.
Write the distance between the ships as a function of time, where T=0
represents 12:00 noon and determine how far apart (in miles) the two ships
will be at 5pm
(1 Knot = 1.15mph)
:
This is a Pythagorus problem: a^2 + b^2 = c^2
Where
a = (5*13*1.15)+80 (travel time is 5 hrs for both ships)
a = 154.75 mi
and
b = 5*11*1.15
b = 63.25 mi
:
c = distance apart in 5 hrs
:
c^2 = 154.75^2 + 63.25^2
c^2 = 23947.5625 + 4000.5625
c^2 = sqrt%2827948.125%29
c = 167.177 mi apart in 5 hrs (5 pm)
:
the function of time equation:
f(t) = sqrt%28%2814.95t%2B80%29%5E2+%2B+%2812.65t%29%5E2%29