Why Velocity is different in different equations?

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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)=-2/3.
Ok, but with the equation of v: v(3)=0.
There are different results! Why?
What am I doing wrong?
 
Last edited:
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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?
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?
 
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?
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
 
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
You are computing two different quantities.
The formula v=4t-12 gives you the velocity at t=3

The formula (distance travelled)/time would give you the average speed.
The formula (change in position)/time would give you the average velocity.
Try it out, by computing how much the traveling object has moved from t=0 to t=3, and then dividing this by 3. Take into account that x(0)≠0.
 
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-13/3 is the average speed from t=0 to t=3 OK thanks a lot.
 
Gjmdp said:
-13/3 is the average speed from t=0 to t=3 OK thanks a lot.
How did you get -13?

Not saying it is wrong, but I find something different:
x(0)=16, x(3)=-2
Change in position=-2-16=-18
Average velocity = -18/3 =-6