I am having trouble with a velocity-versus-time graph. I recently took a Physics test that asked this question: The graph shows the velocity versus time for a particle moving along the $x$ axis. The $x$ position of the particle at $t$=0 seconds was 8 meters. What was the $x$ position of the particle at $t$=2 seconds What was the $x$ position of the particle at $t$=4 seconds?
Here is a link to an image of the graph: http://i.imgur.com/78Hv55D.jpg
So, in order to solve this problem, I used the position equation: $x(t)=x_o + v_ot + \frac{1}{2}at^2$. Starting with the first position, I plugged 2 seconds into $t$, the time; 8 meters into $x_o$, the initial position; 20 meters per second into $v_o$, the initial velocity; and -10 meters per second squared into $a$, the acceleration. I derived the acceleration from the graph using $\frac{\Delta v}{\Delta t}$.
$a = \frac{\Delta v}{\Delta t} = \frac{-20}{2} = -10$
The position equation, with everything in place, reads:
$x(2)=8 + 20(2) + \frac{1}{2}(-10)(2^2)$
This math works out to 28 meters, which is the answer I gave on the test. I did the same thing for the second part of the question--the position of the particle at 4 seconds, as well as the acceleration based on the change in velocity and the change in time.
$a = \frac{\Delta v}{\Delta t} = \frac{-35}{4} = -8.75$
The equation comes out to this:
$x(4)=8+20(4)+\frac{1}{2}(-8.75)(4^2)$
The math comes out to 18 meters. However, according to the curriculum I used, the answer to the first part is 38 meters and the answer to the second part is 16 meters. How can this be the case? Did I fail to apply the position equation correctly? Did I make a mistake when computing the acceleration of the particle? I can't figure out what I did wrong. Is it possible that I used the wrong equation?
I really have no idea. I would appreciate any insight you might have to offer. Thank you in advance for you consideration.
