learn more remarks from the origin to a, the train moves at constant speed in the positive x - direction for…

learn more remarks from the origin to a, the train moves at constant speed in the positive x - direction for the first 4.00 s, because the position vs. time curve is rising steadily toward positive values. from a to b, the train stops at x = 4.00 m for 4.00 s. from b to c, the train travels at increasing speed in the positive x - direction. question what time would it take to produce the displacement represented by a vertical line from 6 m to 8 m in a graph of position vs. time? it would take 0 s. it would take 0.75 s. it would take 2 s. it would take 2.67 s. could an object actually move that distance in that amount of time? yes no exercise hints: getting started | im stuck! use the values from practice it to help you work this exercise. figure (b) above graphs another run of the train. (a) find the average velocity from o to c. if an object travels a net distance of 0, at what average velocity did it travel? m/s (b) find the average and instantaneous velocity from o to a.

learn more remarks from the origin to a, the train moves at constant speed in the positive x - direction for the first 4.00 s, because the position vs. time curve is rising steadily toward positive values. from a to b, the train stops at x = 4.00 m for 4.00 s. from b to c, the train travels at increasing speed in the positive x - direction. question what time would it take to produce the displacement represented by a vertical line from 6 m to 8 m in a graph of position vs. time? it would take 0 s. it would take 0.75 s. it would take 2 s. it would take 2.67 s. could an object actually move that distance in that amount of time? yes no exercise hints: getting started | im stuck! use the values from practice it to help you work this exercise. figure (b) above graphs another run of the train. (a) find the average velocity from o to c. if an object travels a net distance of 0, at what average velocity did it travel? m/s (b) find the average and instantaneous velocity from o to a.

Answer

Explanation:

Step1: Understand position - time graph concept

In a position - time graph, a vertical line represents an instantaneous change in position. Time is plotted on the x - axis and position on the y - axis.

Step2: Determine time for vertical displacement

Since a vertical line in a position - time graph implies no change in the x - value (time), the time taken for the displacement represented by a vertical line is 0 s.

Answer:

It would take 0 s.

For the second part:

Explanation:

An object cannot move a non - zero distance in 0 s as it would imply an infinite velocity which is not physically possible.

Answer:

No

For part (a) of the exercise: The average velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}$. Since the net distance is 0, $\Delta x = 0$. So, $v_{avg}=\frac{0}{\Delta t}=0$ m/s.

Answer:

0 m/s

For part (b), more information about the positions and times at points 0 and A is needed to calculate the average and instantaneous velocities. Since it is not provided completely in the given text, we cannot solve it with the current information.