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

AI Thread Summary
The discussion centers on the incorrect application of the average velocity formula Vavg = (vf + vi)/2 for a non-linear position function x = 12t² - 2t³. The user initially calculated average velocity using this formula and obtained an incorrect result of 9 m/s, while the correct average velocity, calculated using Vavg = Δx/Δt, is 18 m/s. It is clarified that the simpler average formula only applies to linear functions, and for non-linear functions, the average velocity must be derived through integration. The user acknowledges the misunderstanding regarding the conditions for using the simpler formula. Understanding these principles is crucial for accurately calculating average velocity in physics problems.
rainstom07
Messages
15
Reaction score
0
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.
 
Physics news on Phys.org
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?
 
Your average forumla (t0+t3)/2 is only correct if x is a linear function of t ie a straight line.

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

Does this help and can you now obtain the correct answer?
^^ yup
 
Studiot said:
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

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.
 
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.
 

Attachments

  • average1.jpg
    average1.jpg
    9.6 KB · Views: 699
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top