Question

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? 

Part B 

During which trial or trials is the magnitude of the average velocity the largest? 

Answer 1

Part A.

Velocity of an object is given by:

V = dx/dt

dx/dt = slope of graph between x and t

So if in any trial slope of graph is constant, then in that trial velocity will be constant.

From given four trials, trial A and D are straight lines and slope of straight line is always constant, So in these cases velocity of object will be constant.

In trial C object's velocity is zero, since it's position is constant w.r.t to time.

Since Slope of graph in trial B is not constant, So Velocity will also be not constant

Correct option is Trial B.

Part B.

Magnitude of Average velocity is given by:

|Vavg| = |(Xf - Xi)/(tf - ti)|

for trial A, from graph

when t = 0, then x = 0 and when t = 5, then x = 2, So

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

for trial B

when t = 0, then x = -2, when t = 6, then t = 1

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

for trial C

Since object is at same position w.r.t time, So

|Vavg| = 0 m/sec

for trial D, from graph

when t = 0, then x = 4 and when t = 6, then x = -1, So

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

So in trial B and trial D object have maximum magnitude of average velocity.

See that we need magnitude of velocity, So both trial B and D will have highest average velocity.

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