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.

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