# Position time graph describing constant velocity

1. Oct 2, 2007

### cherryrocket

Questions:
a. Q: Find the position at 20s. A: I'm sure it's 56m.

b. Q: Find the maximum displacement
A:
A- 28-0 m = 28m
B- 28-28 m = 0m
C- 56-28m = 28 m
D - 0 - 56m= -56m. This is the one I am worried about, is it correct?

c. Q: Find the velocity in section A of the graph
A: v=m=rise/run = (28-0)m/(8-0)s = 3.5m/s

d. Q: Find the velocity in section D of the graph
A: v=m = rise/run = (0-56)m/(40-20)s = -2.8m/s Is the negative sign correct? Is it negative because the bicycle is going back in the direction of its point of reference?

e. Q: What is happening in section B?
A: The bicycle has remains stopped. No motion.

f. Q: Calculate the average velocity for the whole trip.
A:
A. V= 3.5 m/s
B. V= 0 m/s
C. V= 5.6 m/s
D. V= -2.8 m/s

Vave = (3.5)+(0)+(5.6)+(-2.8) m/s = 6.3 m/s / 4 = 1.575 m/s = 1.58 m/s (significant digits)
Does this seem like an accurate average velocity? Should I have included the result of b which is zero velocity, because using that number greatly reduces my total average velocity...

2. Oct 2, 2007

### cristo

Staff Emeritus
The above is all correct.
I'd use a weighted average here. So, the average velocity = (3.5*8+0*7... etc)/40 (do you see what I'm doing there? You need to weight the average velocity since the four components are not all equal time intervals)

Apart from that, well done!

3. Oct 2, 2007

### Avodyne

For part D, the answer should be +56 m. The maximum displacement is simply the furthest distance the point gets from zero; at a glance, you can see that this is +56m.

Yes, velocity is negative when the object is going backwards.

The easy way to get average velocity is (total displacement)/(total time). In this case, total displacement is zero, so the average velocity is zero. This is equivalent to the "weighted average".