SOLUTION: Spacecraft A is over Houstion at noon on a certain day and traveling at a rate of 275 km/h. Spacecraft B, attemting to overtake and dock with A, is over Houston at 1:15pm and is

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Question 160629This question is from textbook technical mathematics with calculus
: Spacecraft A is over Houstion at noon on a certain day and traveling at a rate of 275 km/h. Spacecraft B, attemting to overtake and dock with A, is over Houston at 1:15pm and is traveling in the same direction as A, at 444 km/h. At what time will B overtake A? At what distance from Houston? This question is from textbook technical mathematics with calculus

Answer by checkley77(12844) About Me  (Show Source):
You can put this solution on YOUR website!
275t=444(t-1.25)
275t=444t-555
275t-444t=-555
-169t=-555
t=-555/-169
t=3.284 hours they will meet.
Proof:
275*3.284=444(3.284-1.25)
903.1=444(2.034)
903.1=903.1