Why can't i apply the simple average velocity formula?

1. Jan 30, 2012

rainstom07

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.

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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.

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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. Jan 30, 2012

Studiot

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?

3. Jan 30, 2012

rainstom07

Thank you! i knew there was some sort of condition attached to the simpler version of average velocity.

^^ yup

4. Jan 30, 2012

willem2

I suppose you meant v_avg = (v_0 + v_3)/2 is only correct if v is a linear function of t.?

X must then be a quadratic function of t, and x = 12t2-2t3 isn't a quadratic.

5. Jan 30, 2012

Studiot

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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