# Homework Help: Why Velocity is different in different equations?

1. Dec 12, 2015

### Gjmdp

1. The problem statement, all variables and given/known data
Calculate velocity. x=position. t=time

2. Relevant equations
x=16-12t+2(t^2)

3. 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: Dec 12, 2015
2. Dec 12, 2015

### Samy_A

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?

3. Dec 12, 2015

### Gjmdp

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

4. Dec 12, 2015

### Samy_A

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 travelling object has moved from t=0 to t=3, and then dividing this by 3. Take into account that x(0)≠0.

Last edited: Dec 12, 2015
5. Dec 12, 2015

### Gjmdp

-13/3 is the average speed from t=0 to t=3 OK thanks a lot.

6. Dec 12, 2015

### Samy_A

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

7. Dec 12, 2015

### Staff: Mentor

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).