# 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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