# Need help with the position of a particle in a velocity-versus-time graph

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?

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.

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I'm not sure why you are using the displacement-time formula if you have the shape of the graph.

The distance covered by the particle is given by the area under the graph from point 0.

After t = 2s, the distance the particle would have covered is the area under the trapezium, that is: $$\dfrac12 (1+2)(20) = 30$$

The position of the particle becomes $8 m + 30m = 38m$.

Maybe you can work out the second part now.

Explanation:

Your formula assumes constant acceleration, which is not the case! For the first 1 s, the acceleration is $0ms^{-2}$, the next 1s, the acceleration is $-20 ms^{-2}$! If you used your formula for $0<t<1$ and then $1<t<2$, then you would have obtained the desired answer. But it's just making the problem more complicated than it actually is at that point :)

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Thank you for clarifying. Very helpful answer. I am confused about something, though. I don't understand why the position equation didn't give me the same result. Is there a reason why I have to use the area as opposed to the equation? Is it because the equation doesn't actually apply to this kind of problem? – ryan4143 Apr 10 '13 at 17:34
@ryan4143 Your formula assumes constant acceleration, which is not the case! For the first 1 s, the acceleration is $0m/s^2$, the next 1s, the acceleration is $-20 m/s^2$! If you used your formula for $0<t<1$ and then $1<t<2$, then you would have obtained the desired answer. But it's just making the problem more complicated than it actually is at that point :) – Jerry Apr 10 '13 at 17:37
Thank you! I really appreciate it. – ryan4143 Apr 10 '13 at 17:38
@ryan4143 You're welcome! – Jerry Apr 10 '13 at 17:38
Hi Jerry - just so you know, our homework policy says not to post complete answers to homework or educational questions, as you did here. Ordinarily we temporarily delete such answers. I'm going to leave this here since it's already been seen and accepted, but please keep the policy in mind for the future. – David Z Apr 10 '13 at 21:32