# Homework Help: Displacment Vs Time

1. Aug 17, 2008

### Gear2d

1. The problem statement, all variables and given/known data

I have a graph where y-axis is displacement, x-axis as time, and rightward movement is positive.

3. The attempt at a solution

A)The graph starts a 0m and moves with a positive slop until it hits 20m (takes 20sec) => This is rightward movement with +v?

B) At 20m for about 20 seconds its a straight line (no slope) => Object is stationary?

C) After 20 seconds (so we are at the 40sec mark on the x-axis) of been stationary it now moves with a slope that is negative until it hits the x-axis at 50seconds where displacement is 0m. => From that stationary point (at 20m) to the 0m the velocity is still positive and that it is now going LEFT? Also why is velocity 2m/s ([0-20]/2) instead of 0m/s, isn't the object back to where it started from?

D)Now from that 0m it still continues until it hits -20m displacement.=> Velocity is negative and is it still moving to the left?

Last edited: Aug 17, 2008
2. Aug 18, 2008

### Kurdt

Staff Emeritus
Seems you have it sorted.

EDIT: Sorry missed your part c question. Since the slope is equal to velocity a negative slope will be a velocity in the direction deemed to be negative, in this case to the left. Yes the object is back where it began but it is still moving with some velocity.

Last edited: Aug 18, 2008
3. Aug 18, 2008

### Gear2d

I still wondering about the velocity where the object goes 20m to the right then 20m back. Shouldn't the velocity be 0m/s, or is that 2m/s the average velocity?

4. Aug 18, 2008

### Kurdt

Staff Emeritus
See the above edit. I think you're confusing velocity with displacement. Velocity is always the slope of the displacement - time graph at that point.

5. Aug 18, 2008

### Gear2d

Thank You Kurdt. I think I see what you are see, somewhat. I was looking at a similar problem (This problem got me confused about why v wasn't 0 on the graph):

I have point A and point B:

A..........................................................B (straight line, 10m apart) If a person walks from A to B and back to A in 2 seconds, what is the velocity?

From this I see that he walked 10m right then 10 left so displacement is 0, so v=displacement/time = 0.

So I was trying to relate this problem to that of the displacement vs. time graph where the ball went 20m right then 20m back.

So when you said velocity is slope of Disp/time, you are saying is that at that 20m mark to the 0m is independent from the starting point where the ball went 0m to 20m. So the slope, where I confusing the velocity (20m back to 0m), is the velocity of ball going back. So the only time v=0 (from 20m back to 0m) is at the 0m mark. So before that the ball had a +v(going left) and AFTER it passes the 0m mark has a -v (stilling going left).

Last edited: Aug 18, 2008
6. Aug 18, 2008

### Kurdt

Staff Emeritus
The slope is always negative indicating a velocity in the left direction. Average velocity is the displacement divided by time. The instantaneous velocity is different and is the rate of change of displacement. I hope thats a bit clearer.