Is The Velocity Of This Object Increasing Or Decreasing?

[kingandsona]

Homework Statement

Hi everyone,
8th grade physics teacher here.

We are doing a handout where we have to identify time-velocity graphs with descriptions. Can you help me with this one?

https://imgur.com/a/vEIkg

A) The velocity is increasing at a uniform rate.
B) The velocity is decreasing at a uniform rate.

Homework Equations

The Attempt at a Solution

At first I thought it was decreasing but then an 8th grade student said 1) the slope is positive and 2) if you continue the line after the x-axis it will be increasing. Also, the Vf is literally higher on the y-axis than Vi.dd

I think the same question could be raised about displacement that is negative but with a positive slope approaching the x-axis. The displacement on the y-axis would be going up, but it is decreasing since it's getting closer to the starting point.

Let me know your thoughts!

 

[BvU]

You want to clearly distinguish "is positive/negative" from "is increasing/decreasing"

Positive velocity means "is moving in the positive direction"
Negative velocity means "is moving in the negative direction" (backwards, if you want)

The plot describes a velocity that is negative at the start but becomes more and more positive as time proceeds. The straight line shows that it increases at a uniform rate (+1 unit per +2 time units)

Ah, and: hello king&sonA, . Happy teaching !

 

[Ray Vickson]

Homework Statement

Hi everyone,
8th grade physics teacher here.

We are doing a handout where we have to identify time-velocity graphs with descriptions. Can you help me with this one?

https://imgur.com/a/vEIkg

A) The velocity is increasing at a uniform rate.
B) The velocity is decreasing at a uniform rate.

Homework Equations

The Attempt at a Solution

At first I thought it was decreasing but then an 8th grade student said 1) the slope is positive and 2) if you continue the line after the x-axis it will be increasing. Also, the Vf is literally higher on the y-axis than Vi.dd

I think the same question could be raised about displacement that is negative but with a positive slope approaching the x-axis. The displacement on the y-axis would be going up, but it is decreasing since it's getting closer to the starting point.

Let me know your thoughts!

Velocity can be positive or negative, and is negative in your graph. However, speed is the magnitude of velocity, so is always >= 0.

In your case, velocity is increasing and speed is decreasing.

 

[hilbert2]

In the case of 2d or 3d motion there would be no "positive" and "negative" directions, so in that case any motion with decreasing speed could also be described as having a decreasing velocity.

 

[NoahCygnus]

Look at the graph of again, think of it this way. First the car is moving in the backward direction as the velocity is negative, but then it's slowing down as the velocity in the negative direction is decreasing in magnitude and then comes to a halt after some time. The negative sign just tells you the direction of motion in 1D. Why are you looking at the slope? This is not position time graph . The slope of velocity time graph gives you acceleration and the acceleration in this case is constant.

 

[Ray Vickson]

Look at the graph of again, think of it this way. First the car is moving in the backward direction as the velocity is negative, but then it's slowing down as the velocity in the negative direction is decreasing in magnitude and then comes to a halt after some time. The negative sign just tells you the direction of motion in 1D. Why are you looking at the slope? This is not position time graph . The slope of velocity time graph gives you acceleration and the acceleration in this case is constant.

He is looking at the slope because that tells him the rate of increase or decrease in velocity

 

[NoahCygnus]

He is looking at the slope because that tells him the rate of increase or decrease in velocity

Well in that case the acceleration is a constant and positive, the velocity is negative so the velocity will decrease and hit zero and then start increasing at a constant rate in the positive direction.

 

[haruspex]

Well in that case the acceleration is a constant and positive, the velocity is negative so the velocity will decrease

No, the magnitude of the velocity (i.e. the speed) will decrease, but the velocity becomes less negative, so is increasing.

 

[NoahCygnus]

No, the magnitude of the velocity (i.e. the speed) will decrease, but the velocity becomes less negative, so is increasing.

That is what I said in my first post.

the velocity in the negative direction is decreasing in magnitude

 

[haruspex]

That is what I said in my first post.

But in post #7 you wrote

the velocity will decrease

which is incorrect.
Besides "velocity in the negative direction" doesn't really mean anything. By definition, velocity is measured in the positive direction.

 

[NoahCygnus]

But in post #7 you wrote

which is incorrect.

Will decrease in magnitude in the negative direction. I guess it wasn't self explanatory that by saying "Velocity will decrease", I meant it will decrease it's magnitude in the negative direction and hit zero and then increase its magnitude again but the direction of the velocity vector will be the positive direction.

 

[haruspex]

I guess it wasn't self explanatory that by saying "Velocity will decrease"

Whatever you intended, the statement "Velocity will decrease" is simply wrong. These are well-defined terms. If you meant something else then you used the wrong terminology.

 

[Ray Vickson]

Well in that case the acceleration is a constant and positive, the velocity is negative so the velocity will decrease and hit zero and then start increasing at a constant rate in the positive direction.

You cannot say that because you have no way of knowing what the graph looks like outside the region plotted.

 

[PeroK]

What you all missed is that this post is identical, word for word, to the one below from the 19th of December.

 

[jbriggs444]

Why so it is. Different author names, but identical post. How droll.

 

[BvU]

Look at the graph of again, think of it this way. First the car is moving in the backward direction ...

Funny you bring in the word "car" -- I didn't see it anywhere

 

[NoahCygnus]

Funny you bring in the word "car" -- I didn't see it anywhere

So just because it's not given in the problem I can't assume the particle undergoing motion is a car. All right.

 

[CWatters]

It might seem strange but the direction you define as "positive" determines if the velocity is increasing or decreasing.