# Help with total distance problem

1. Sep 12, 2004

### ramin86

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.

Last edited: Sep 12, 2004
2. Sep 12, 2004

### Pyrrhus

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.

3. Sep 12, 2004

### ramin86

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.

4. Sep 12, 2004

### Pyrrhus

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:

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

$$x = 100 - 37.5 = 62.5 m$$

$$x_{total} = (100 +100 + 62.5) m = 262.5 m$$

5. Sep 12, 2004

### ramin86

Wait, I thought x final = Xi + Vxi*t + 1/2*a*t^2

6. Sep 12, 2004

### Pyrrhus

It is i just simplified.