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Why can't i apply the simple average velocity formula? |
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| Jan30-12, 04:06 PM | #1 |
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Why can't i apply the simple average velocity formula?
I'm doing a homework problem (i already know the answer) and i came across an error in my logic/application of the formula Vavg = (v + v0)/2. Hopefully you can help me understand why it's incorrect to use the formula.
x = 12t2-2t3 describes a particle position. the derivative of x = 24t-6t2. The homework question asked me find the average velocity between t = 0 and t = 3. Using the formula Vavg = Δx/Δt yields 18 m/s... the correct answer. --- When i use the simpler formula: Vavg = (vf + vi) / 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. |
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| Jan30-12, 04:31 PM | #2 |
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Hello rainstom, welcome to Physics Forums
Your average forumla (t0+t3)/2 is only correct if x is a linear function of t ie a straight line. The function x = 12t2-2t3 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? |
| Jan30-12, 04:42 PM | #3 |
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| Jan30-12, 05:58 PM | #4 |
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Why can't i apply the simple average velocity formula?X must then be a quadratic function of t, and x = 12t2-2t3 isn't a quadratic. |
| Jan30-12, 06:31 PM | #5 |
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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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