Question 215507This question is from textbook
: My question is a precalculus question and it wants us to basically translate word problem into a function
Car A passes point O heading east at a constant rate of 40mi/h; car B passes the same point 1 hour later heading north at a constant rate of 60mi/h. Express the distance between the cars as a function of time t, where t is measured starting when car B passes point O.
I drew a diagram of this, and thought of using the Pythagorean Theorum to find the distance between car A and Car b, but I just don't know how to incoporate the "1 hour later part." the answer in the book is 20
This question is from textbook
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Car A passes point O heading east at a constant rate of 40mi/h; car B passes the same point 1 hour later heading north at a constant rate of 60mi/h.
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Car A DATA:
rate = 40 mph ; time = t+1 hrs ; distance = rt = 40(t+1) miles
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Car B DATA:
rate = 60 mph ; time = t hrs ; distance = 60t miles
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Express the distance between the cars as a function of time t, where t is measured starting when car B passes point O.
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Yes, use the Pythagorean Equation to get:
(distance between A and B) = sqrt[(60t)^2 + (40(t+1))^2]
d = 3600t^2 + 1600(t^2+2t+1)
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Cheers,
Stan H.
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