Quantitatively Compare the values of objects A and B

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In summary, the conversation was discussing how to solve for average velocity and displacement using the attached graph. The equations for average velocity and displacement were provided, and it was determined that for object B, the average velocity was 6m/s and the displacement was 10m from t = 0s to t = 2.5s. For object A, the average velocity was 4m/s and the displacement was 10m from t = 0s to t = 2.5s. It was also determined that for both objects, the average velocity was 4m/s from t = 0s to t = 5s, and the displacement was 20m from t = 0s to t =
  • #1

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
[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.
  1. 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.
  2. [itex]\stackrel{8+6+4+2+0}{5}[/itex] = 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.
 

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  • #2
AnthonyTufo said:

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
[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.
  1. 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.
  2. [itex]\stackrel{8+6+4+2+0}{5}[/itex] = 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 .

Vavg = (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 Vavg=(V1+V2)/2 .

Approach 2) looks to be simpler in this case.
 
  • #3
Tanya Sharma said:
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 .

Vavg = (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 Vavg=(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
 

1. What does it mean to quantitatively compare the values of objects A and B?

Quantitatively comparing the values of objects A and B means to use numerical data or measurements to determine the similarities or differences between the two objects. This involves using mathematical calculations and statistical analysis to make a quantitative comparison.

2. How is the quantitative comparison of objects A and B different from a qualitative comparison?

A quantitative comparison focuses on numerical data and measurements, while a qualitative comparison focuses on qualities or characteristics of the objects. Quantitative comparisons provide more precise and measurable results, while qualitative comparisons are more subjective.

3. What are some methods used to quantitatively compare the values of objects A and B?

Some methods for quantitatively comparing the values of objects A and B include statistical analysis, mathematical calculations, and data visualization techniques such as graphs and charts. These methods help to identify patterns, trends, and relationships between the two objects.

4. Can the values of objects A and B be compared using only one measurement or data point?

No, it is not recommended to compare the values of objects A and B using only one measurement or data point. It is important to use multiple measurements and data points to get a more accurate and comprehensive understanding of the comparison between the two objects.

5. How can a quantitative comparison of objects A and B be useful in a scientific study?

A quantitative comparison can be useful in a scientific study as it provides objective and measurable data to support or refute a hypothesis. It also allows for a more precise and accurate understanding of the relationship between the two objects, which can aid in making informed conclusions and decisions.

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