Here is the question:

 

Displacement for an object in metres, x, is given by the equation:

x=1.5t^2 + 4t + 2

 

Find the average velocity from t=2 to t=4.

 

 

Now, every single person in the class, including the teacher, did this method:

t=2: 1.5(2)^2 + 4(2) + 2 = 16

t=4: 1.5(4)^2 + 4(4) + 4 = 42

(42-16)/(4-2) = 13

 

 

My method was this:

t=2: 1.5(2)^2 + 4(2) + 2 = 16m over 2 seconds, or 8 m/s

t=4: 1.5(4)^2 + 4(4) + 4 = 42m over 4 seconds, or 10.5 m/s

Therefore, the average velocity from t=2 to t=4 would be:

(8+10.5)/(2) or 9.25 m/s

 

Essentially, the first method gives the average DISPLACEMENT from 2 to 4, whereas my method is giving the average VELOCITY.

 

Which method is correct?

ooh. cant remember how to do these simple questions.

 

dx/dt = 3t + 4

 

since its linear. the average velocity is [3(4) + 4 - 3(2) - 4]/2 = 3

 

i may be wrong

Never mind, I was corrected by a friend. Right method with the derivation, but the correct formula would be [(3(2)+4) + (3(4)+4)] / 2 or 13.

oh yeah . sorry about that. messed up my plus and minuses.

 

as you said. its just that