SOLUTION: Spacecraft A is over Houston at noon on a certain day and travelling at a rate of 300Km/h . Spacecraft B , attempting to overtake and dock with A , is over Houston at 1:15 P.M. a

Question 1055391: Spacecraft A is over Houston at noon on a certain day and travelling at a rate of 300Km/h . Spacecraft B , attempting to overtake and dock with A , is over Houston at 1:15 P.M. and is traveling in the same direction as , at A . 410 km/h At what time will B overtake A ? At what distance from Houston?
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
This can be treated as if both crafts departed from "Houston", at their described times and speeds (and in the same direction). Deal with time quantities instead of time points. This is a typical two-traveler catchup problem.

                 SPEED      TIME         DISTANCE

A                 300                d

B                 410        t              d

Can you form the equation or equations necessary to solve for t and use it to determine the time?

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A start in case of difficulty:
Travel Rate Rule for Constant Rate:
RT=D, or D/T=R for constant R.

The catchup distance, d, for craft B to travel, is the same value, both crafts. Think of the starting position point as being Houston. A travels for longer time than does B. Be sure this makes sense to you; otherwise, you will struggle to understand this next step:

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Next possible 3 or 4 steps:

------continue from here.

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I found the time QUANTITY to be and the time when, to be 7:36PM.
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