SOLUTION: A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour. Another ship leaves
the same port at 1:00 PM and sails North at a speed of 20 miles/hour. At what time
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Question 746316: A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour. Another ship leaves
the same port at 1:00 PM and sails North at a speed of 20 miles/hour. At what time are the two
ships going to be 50 miles apart from each other? (Hint: Distance = Speed * Time)
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
A ship leaves a port at 12:00 Noon and sails East at speed of 10 miles/hour.
Another ship leaves the same port at 1:00 PM and sails North at a speed of 20 miles/hour.
At what time are the two ships going to be 50 miles apart from each other?
:
Let t = sailing time of the Northbound ship (From 1 PM)
then
(t+1) = sailing time of the Eastbound
:
This is a pythag problem a^2 + b^2 = c^2; where:
a = 20t
b = 10(t+1)
c = 50 mi
(20t)^2 + (10t+10)^2 = 50^2
400t^2 + 100t^2 + 200t + 100 = 2500
A quadratic equation
500t^2 + 200t + 100 - 2500 = 0
500t^2 + 200t - 2400 = 0
simplify, divide by 100
5t^2 + 2t - 24 = 0
Factors to
(5t+12)(t-2) = 0
the positive solution
t = 2 hrs, therefore at 3 PM, they will be 50 mi apart
:
:
Check this out
2(20) = 40 mi traveled by northbound ship
3(10) = 30 mi traveled by the eastbound
Enter into your calc; enter: = 50
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