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Home/ Questions/An object moves along the x axis during four separate trials.
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lyytutoria
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lyytutoriaExpert
Asked: April 11, 20222022-04-11T18:45:14+00:00 2022-04-11T18:45:14+00:00In: Physics

An object moves along the x axis during four separate trials.

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Velocity from Graphs of Position versus Time Review PartA An object moves along the x axis...

Velocity from Graphs of Position versus Time 
An object moves along the x axis during four separate trials. Graphs of position versus time for each trial (with the same scales on each axis) are shown in the figure. (Figure 1)
Part A
During which trial or trials is the object’s velocity not constant?
Check all that apply.
1, Trial A 
2, Trial B
3, Trial C
4, Trial D
Part B
During which trial or trials is the magnitude of the average velocity the largest? 
Check all that apply.
1, Trial A
2, Trial B
3, Trial C
4, Trial D

♦ Relevant knowledge

Velocity is an inverse vector that has both a magnitude and direction. It determines the distance of an object in unit of time. The term “speed” is the amount of distance an object travels during a specific period of time.
magnitudetrialvelocity
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    lyytutoria Expert
    2022-04-24T15:35:27+00:00Added an answer on April 24, 2022 at 3:35 pm

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  2. Susan O'Sullivan
    2022-04-11T18:45:17+00:00Added an answer on April 11, 2022 at 6:45 pm

    Part A.

    The following factors determine the velocity of an object:

    V = dx/dt

    dx/dt = slope between x and the graph.

    If the trial slope of a graph is constant, then that trial velocity will also be constant.

    Given four trials, A and D are straightlines. The slope of straight lines is constant so velocity of object in these cases will also be constant.

    Trial C object’s velocity in trial is zero because its position is constant with time.

    Because Slope of graph in trial A is not constant, Velocity will not be constant.

    The correct option is Trial B.

    Part B.

    The Magnitude of the Average velocity is given below:

    |Vavg| = |(Xf – Xi)/(tf – ti)|

    Trial A: graph

    If t is 0, then x equals 0 and t=5, then x equals 2.

    |Vavg| = |(2 – 0)/(5 – 0)| = 0.4 m/sec

    For trial B

    If t is 0, then x equals -2. When t is 6, then t equals 1.

    |Vavg| = |(1 – (-2))/(6 – 0)| = 0.5 m/sec

    For trial C

    Because object is in the same place at all times, So

    |Vavg| = 0 m/sec

    Trial D, graph

    If t is 0, then x equals 4 and t=6, then x equals 1.

    |Vavg| = |(-1 – 2)/(6 – 0)| = 0.5 m/sec

    Thus, in trials B and D, the maximum velocity is achieved for both objects.

    We need velocity magnitude. So trial A and D will both have the highest average velocity.

    Please upvote.

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