
#1
Jan3012, 04:06 PM

P: 16

I'm doing a homework problem (i already know the answer) and i came across an error in my logic/application of the formula V_{avg} = (v + v_{0})/2. Hopefully you can help me understand why it's incorrect to use the formula.
x = 12t^{2}2t^{3} describes a particle position. the derivative of x = 24t6t^{2}. The homework question asked me find the average velocity between t = 0 and t = 3. Using the formula V_{avg} = Δx/Δt yields 18 m/s... the correct answer.  When i use the simpler formula: V_{avg} = (v_{f} + v_{i}) / 2 = (x'(3.0)+x'(0.0))/2 = 18/2. I get 9 m/s which is incorrect.  Adding the velocity at t=3 with the velocity at t=0 and then dividing by 2 should've produced 18 m/s... My logic is clearly wrong, but how? x' describes the velocity of the particle at t seconds? right? thanks. 



#2
Jan3012, 04:31 PM

P: 5,462

Hello rainstom, welcome to Physics Forums
Your average forumla (t_{0}+t_{3})/2 is only correct if x is a linear function of t ie a straight line. The function x = 12t^{2}2t^{3} is decidedly non linear The way to derive an average for a non linear function (works for linear as well but is trivial) is to integrate the function and divide by the interval or number of points or samples. So average = Area/Interval Does this help and can you now obtain the correct answer? 



#3
Jan3012, 04:42 PM

P: 16





#4
Jan3012, 05:58 PM

P: 1,351

Why can't i apply the simple average velocity formula?X must then be a quadratic function of t, and x = 12t^{2}2t^{3} isn't a quadratic. 



#5
Jan3012, 06:31 PM

P: 5,462

Hello willem2 does this attachment help?
The average velocity is the number which if you multiplied it by the time would give you the total distance travelled. 


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