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Say I have an object that starts falling from rest at 2 m/s2. Its total displacement and velocity at the end of every second should look something like this:

t    d    v
1    2    2
2    6    4
3    12   6
4    20   8

At the end of the first second, it should have accelerated to 2 m/s and moved 2 m. At the end of 4 seconds, it should moved 20 m. But when I use the equation for getting displacement, $d=\frac{1}{2}at^2$, I get 16 m. Shouldn't it be 20 m? What is going wrong here?

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1  
With $t=1$ and $a=2$, shouldn't $d(1) = 0.5*2*(1*1) = 1$?? –  Kyle Kanos Aug 6 '13 at 1:01
    
@KyleKanos I guess, but even that doesn't make sense. It should have moved 2 meters, according to that it accelerates at 2 m/s^2 –  Dan the Man Aug 6 '13 at 1:05
2  
If you took the time to enter $t$ into $d=at^2/2$, you would see that your entire d column is wrong. So the question really is: what are you using to compute $d$? –  Kyle Kanos Aug 6 '13 at 1:08
    
@KyleKanos I didn't realize that for the first second it should have been 1m. I then added what should have corresponded for the next seconds. –  Dan the Man Aug 6 '13 at 1:12
1  
I echo what @Kyle said. Without some explanation of why you expect $d$ at 1s to be 2m, I really don't think this is a good question. It comes across as "why don't these numbers I made up fit this formula?" So could you edit the question to explain where you're getting your numbers from? –  David Z Aug 6 '13 at 1:27

1 Answer 1

up vote 4 down vote accepted

What it looks like you are doing is calculating the speed at the end of the interval, then adding that to the distance travelled. As pointed out, that is incorrect. The distance travelled over an interval is the average speed times the length of the interval. In this case, since the acceleration is constant, the average speed is just the average of the initial and final speed of the interval.

t.   v     v_avg       d
0    0      0          0
1    2      1          1
2    4      3          4
3    6      5          9
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