- #1

AnthonyTufo

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## Homework Statement

*See attached graph*

If A has the greater value answer A.

If B has the greater value answer B.

if equal answer equal.

- Average Velocity from t = 0s to t =2.5s
- Average Velocity from t = 0s to t =5s
- Displacement from t = 0s to t =2.5s
- Displacement from t = 0s to t = 5s

## Homework Equations

__Average Velocity__

[itex]\overline{v}[/itex]: average velocity

S: sum of numbers

N: the amount of terms

[itex]\overline{v}[/itex] = S/N

__Displacement__

l = length

w = width

for B, Displacement = .5lw

for A, Displacement = lw

## The Attempt at a Solution

I got all these wrong on the test, but I think I understand both the average velocity problems now.

- For B: [itex]\stackrel{8+6+4}{3}[/itex] = 18/3 = 6m/s and A is always 4m/s, so the answer is
__B is greater than A__. Or I could have just looked at the graph and noticed that object B's line is above A's.

- [itex]\stackrel{8+6+4+2+0}{5}[/itex] = 20/5 = 4m/s answer is the
__values are equal.__ - Displacement of B = .5 (8m/s) (2.5s) = 10m

Displacement of A = (4m/s) (2.5s) = 10m

I said equal and that was wrong, I think I am approaching this incorrectly. - Displacement of B = .5 (8m/s) (5s) = 20m

Displacement of A = (4m/s) (5s) = 20m

For this one I originally thought adding the displacements of B together would work (10m + (.5) (4m/s)(2.5s) = 15m), but that doesn't follow order of operations(and it leaves out a triangular space when I look at it graphically), right? I don't understand this though. Shouldn't the displacement be getting smaller as the velocity decreases? I can see it graphically, but I don't understand it fundamentally.__This answer is they are equal.__

__B as travelling a total of 15m__and

__A travelling a total 20m__and they were both marked correct, so there may be some confusion on my part in assuming my teacher didn't make a mistake.

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