SOLUTION: A ship leaves Port Canaveral on May 6th @ 1200GMT bound for Gibraltar, a distance of 3822 miles. They expect to average 13.5 kts and give an estimated time of arrival ( ETA ) of M

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: A ship leaves Port Canaveral on May 6th @ 1200GMT bound for Gibraltar, a distance of 3822 miles. They expect to average 13.5 kts and give an estimated time of arrival ( ETA ) of M      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


   



Question 446153: A ship leaves Port Canaveral on May 6th @ 1200GMT bound for Gibraltar, a distance of 3822 miles. They expect to average 13.5 kts and give an estimated time of arrival ( ETA ) of May 18th at 0700 GMT. On the 13th their position at 0100 shows they have averaged 14.4 and have 1554 miles left to go, they are ahead of schedule. Due to limitations with the engines they can not slow to less than 10 knots. If they continue to average 14.4kts, When will they be able to slow to 10kts and still arrive at the original ETA of 0700 on the 18th?
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
dep 6th May 1200
Arr 18th may 7.00
Time = 11 days + 19 hours = 283 hours
scheduled speed = 13.5 kts
distance = 3822 miles
..
13th May 0100 hours time elapsed =6 days and 13 hours = 157 hours
To reach on schedule it has 283-157 = 126 hours
Let it travel x kts at 14.4 kts/h
time = x/14.4
Balance at 10 kts/h
Time at 10 kts/hr =(1554-x)/10
x/14.4 + (1554-x)/10= 126
10x+14.4(1554-x)= 10*14.4*126
10x+22,377.6 -14.4x =18144
-4.4x=-4233.6
x=962.18 kts.
At 14.4 kts/hr it will take 962.18/14.4 ---hours
=66.81 hours ===> 2 days 18 hours 48 minutes
16th may 1948 hours it has to reduce to 10 kts/hr to be on schedule.