The discussion revolves around the calculation of velocity from a given position function, specifically the equation x=16-12t+2(t^2). Participants are exploring the differences in velocity results obtained from the derivative of the position function and from average velocity calculations over a specified time interval.
Some participants have provided insights into the differences between instantaneous velocity and average velocity, suggesting that the original poster may be conflating these concepts. There is ongoing exploration of the definitions and calculations involved, with no explicit consensus reached.
Participants note that the initial position at t=0 is not zero, which affects the average velocity calculation. There is also mention of the need to consider the change in position over time when calculating average velocity.
Using your equations: x(3)=16-12*3+2*9=16-36+18=-2 and v(3)=0.Gjmdp said:Homework Statement
Calculate velocity. x=position. t=time
Homework Equations
x=16-12t+2(t^2)
The Attempt at a Solution
Derivative of x: v=4t-12
Ok let's try:
Equation of x: x=16-12t+2(t^2)
Equation of v: v=4t-12
With the equation of x: x(3)=2 then v(3)=3/2.
Ok, but with the equation of v: v(3)=0.
There are different results! Why?
What am I doing wrong?
So x(3)=-2 in t=3.Samy_A said:Using your equations: x(3)=16-12*3+2*9=16-36+18=-2 and v(3)=0.
How do you get v(3)=3/2?
You are computing two different quantities.Gjmdp said:So x(3)=-2 in t=3.
v=x/t
v(3)=-2/3
I was wrong.
But it keep being different
v=0: v=-2/3
How did you get -13?Gjmdp said:-13/3 is the average speed from t=0 to t=3 OK thanks a lot.
v is not equal to x/t. This is only correct if x =0 at t = 0 (which in your problem it is not) and if v is constant (which in your problem it is not).Gjmdp said:So x(3)=-2 in t=3.
v=x/t
v(3)=-2/3
Sorry, I was wrong.
But it keep being different
v(3)=0: v(3)=-2/3