Why Velocity is different in different equations?

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Velocity can be calculated using the derivative of the position function, resulting in v=4t-12. When evaluating at t=3, different methods yield varying results: using the derivative gives v(3)=0, while calculating average velocity from position changes gives -6. The discrepancy arises because the average velocity considers the total change in position over time, while the derivative provides instantaneous velocity. It's crucial to differentiate between average velocity and instantaneous velocity, especially when initial conditions are not zero. Understanding these concepts clarifies why different equations yield different velocity results.
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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.
 
Last edited:
-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
 
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
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).
 
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