No derivatives or Calculus needed here, it is a basic algebra problem.
I assume they left at the same time,
so they travelled the same time when they pass
distance covered by Beth = 60t
distance covered by Alving = 20t+40
that difference is 30 miles
20t + 40 - 60t = 30
-40t = -10
t = 1/4
They pass each other after 15 minutes
so distance that Beth went is (1/4)(60) = 15 miles
So they pass each other 15 miles north of Holland
Beth leaves Muskegon, 30 mile north of Holland, traveling at 60 mph.
Alvin leaves Holland traveling north at V=20t+40 mi/hr.
When will Alvin pass Beth?
How far from Holland will they be?
:I think Beth's distance is d=60t-30 (using Holland as the frame of reference). Should I just take the anti-derivative of Alvin's velocity and then set those functions equal to each other and solve for time? Thanks.
2 answers
you are wrong.
the answer is 3 hours and was achieved by taking the anti-derivative like I said.
Beth's distance is 60t+30.
To get Alvin's distance you must take the anti-derivative (10t^2 +40t +C) C=0.
Then set the equations equal to each other and solve for t. t= 3, -1
Your answer did not make sense. They pass each other 210 mile away from Holland.
the answer is 3 hours and was achieved by taking the anti-derivative like I said.
Beth's distance is 60t+30.
To get Alvin's distance you must take the anti-derivative (10t^2 +40t +C) C=0.
Then set the equations equal to each other and solve for t. t= 3, -1
Your answer did not make sense. They pass each other 210 mile away from Holland.
1 answer