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anonymous
 5 years ago
It takes me 1.5 hours to go to the college and 2 hours to come back. I fmy average speed while going to the college is 2mph faster than my speed while comming back, find my speed while going to the college.
anonymous
 5 years ago
It takes me 1.5 hours to go to the college and 2 hours to come back. I fmy average speed while going to the college is 2mph faster than my speed while comming back, find my speed while going to the college.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S1 = SPEED TO COLLEGE, S2 = SPEED FROM COLLEGE, D = DISTANCE D(S1 + S2) = 1.5 S1 = S2 + 2 D[(S2 + 2) + S2] = 1.5 2(S2) + 2 = 1.5/D UNABLE TO SOLVE WITHOUT THE DISTANCE TO COLLEGE, HOWEVER BY USING THE DERIVATIVE OF 2(S2) + 2  1.5D^1 = 2 + (1.5D)^2 YOU CAN CALCULATE THE RATE OF CHANGE BETWEEN THE DISTANCE AND SPEED

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You can calculate average velocity with v = dt. You're given the t1 and t2 as 2 and 1.5 respectively where t1 is the first trip and t2 is the 2nd trip. You're also given your v1 in terms of your v2. When it says v1 is 2 mph faster than v2, it's telling you v1 = v2 +2, right? The distance is the same in each one, so you can leave it as just d. You can then set it up as a series of equations. v1=t1d and v2 = t1d. Use your v1 = v2 + 2 to solve the distance from there. Once you have your distance, it becomes easy to solve for your velocities.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, v2 = t2d, not v2 = t1d
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