Quantitatively Compare the values of objects A and B

[AnthonyTufo]

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.

1. Average Velocity from t = 0s to t =2.5s
2. Average Velocity from t = 0s to t =5s
3. Displacement from t = 0s to t =2.5s
4. Displacement from t = 0s to t = 5s

Homework Equations

Average Velocity
\overline{v}: average velocity
S: sum of numbers
N: the amount of terms
\overline{v} = 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.

1. For B: \stackrel{8+6+4}{3} = 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.
   
2. \stackrel{8+6+4+2+0}{5} = 20/5 = 4m/s answer is the values are equal.
3. 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.
4. 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.

I also had to make a motion map and on that I put B as traveling a total of 15m and A traveling 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.

 

[Tanya Sharma]

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.

1. Average Velocity from t = 0s to t =2.5s
2. Average Velocity from t = 0s to t =5s
3. Displacement from t = 0s to t =2.5s
4. Displacement from t = 0s to t = 5s

Homework Equations

Average Velocity
\overline{v}: average velocity
S: sum of numbers
N: the amount of terms
\overline{v} = 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.

1. For B: \stackrel{8+6+4}{3} = 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.
   
2. \stackrel{8+6+4+2+0}{5} = 20/5 = 4m/s answer is the values are equal.
3. 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.
4. 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.

I also had to make a motion map and on that I put B as traveling a total of 15m and A traveling 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.

Hi AnthonyTufo

Welcome to Physicsforums!

There are two ways you can think about this problem.

1)You can calculate displacement by calculating area under the v-t graph .

V_avg = (X₂-X₁)/(t₂-t₁) = ΔX/Δt,where X₂ and X₁ are displacement of the object at time t₂ and t₁ respectively .

2)For the special case of straight line motion ,if the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as V_avg=(V₁+V₂)/2 .

Approach 2) looks to be simpler in this case.

 

[AnthonyTufo]

Hi AnthonyTufo

Welcome to Physicsforums!

There are two ways you can think about this problem.

1)You can calculate displacement by calculating area under the v-t graph .

V_avg = (X₂-X₁)/(t₂-t₁) = ΔX/Δt,where X₂ and X₁ are displacement of the object at time t₂ and t₁ respectively .

2)For the special case of straight line motion ,if the beginning and ending velocities for this motion are known, and the acceleration is constant, the average velocity can also be expressed as V_avg=(V₁+V₂)/2 .

Approach 2) looks to be simpler in this case.

Am I still wrong for problems 1 and 2? If not, I still need help on verifying 4 and figuring out 3.
Okay, I see what I did wrong. Too find displacement I should have used the ENTIRE time shown on the graph and because it is constant this is easy. I found this: http://www.physicsclassroom.com/class/1dkin/u1l4e.cfm