Help with total distance problem

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The discussion revolves around calculating the total distance traveled over 20 seconds using a provided chart. The user initially calculated speeds at 10 seconds and 20 seconds as 20 m/s and 5 m/s, respectively. They attempted to find the total distance by breaking it into segments but encountered errors in their calculations, leading to confusion about the correct method. After several attempts, they derived a total distance of 262.5 meters using different equations of motion. The conversation highlights the importance of correctly applying kinematic equations and understanding the effects of acceleration on distance calculations.
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I was given the following chart, http://www.webassign.net/pse/p2-14.gif, it asked to find the speed at 10s, and 20s, I got 20m/s and 5m/s respectively. The last part asks to find the total distance traveled in the first 20 seconds. Now I did x final = x initial + 1/2(Vxi + Vxf)t and got 250 m, I wanted to know if this looks right, since I have only one more chance to answer the question.
 
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Look! the acceleration is constant from 0 s to 10 s, 10 s to 15 s, and 15 s to 20 s. Just calculate the distance per piece and add it up for total.
 
I tried that, I did 20 m/s * 10, plus 20 m/s * 5, because the acceleration is zero for those 5 seconds, plus 5 m/s * 5, which came out to total of 325, but it turned out to be wrong. I also tried replacing 5 m/s *5 with -15 m/s * 5 (velocity for that segment alone), and that also turned out to be wrong.
 
Well for the first part:

V = 2t

V = 20 m/s

x = \frac{1}{2}2t^2

x = 100 m

for the second part:

x = v_{o}t

x = 100 m

for the third part:

x = 20t - \frac{1}{2}3t^2

x = 100 - 37.5 = 62.5 m

Adding:

x_{total} = (100 +100 + 62.5) m = 262.5 m
 
Wait, I thought x final = Xi + Vxi*t + 1/2*a*t^2
 
It is i just simplified.
 
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