What is the Average Velocity and Speed of a Moving Ball?

[Heat]

Homework Statement

http://img2.freeimagehosting.net/uploads/2256719644.jpg​

A ball moves in a straight line (the x-axis). The graph in the figure shows this ball's velocity as a function of time.

What are the ball's average velocity during the first 3.0s ?

What are the ball's average speed during the first 3.0s ?

Suppose that the ball moved in such a way that the graph segment after 2.0s was -3.0 m/s instead of +3.0m/s . Find the ball's and average velocity during the first 3.0s in this case.

Suppose that the ball moved in such a way that the graph segment after 2.0s was -3.0m/s instead of +3.0 m/s. Find the ball's average speed during the first 3.0s in this case.

The Attempt at a Solution

Ok, from my reading I understand that average speed is not the magnitude of average velocity.
To determine average speed you need total total distance/ total time.
I know that if the displacement is 0, then the average velocity is 0.


What I have done so far is get the slop from time 0 through time =3s. It gives me 1.
Since, having the tangent line slop to the three seconds, would be instantaneous speed, then I would not assume that speed = 0.

My other try is that I start at the point from 0s @ 2m/s through 3s @ 3m/s.
The line would have a slope of 1/3 resulting in the average speed of .66m/s. Is this right, if not can you point me in the right direction.

 

[berkeman]

I'm not sure I understand the original problem statement (part is missing?), but keep in mind that speed is a scalar, and velocity is a vector. So the average speed is not necessarily the total distance divided by the total time (that would be the average velocity, with an appropriate direction). If you drive one way at 60kph, turn around and drive back at 30kph for twice as long, the average speed is not zero...

 

[learningphysics]

I'm not sure I understand the original problem statement (part is missing?), but keep in mind that speed is a scalar, and velocity is a vector. So the average speed is not necessarily the total distance divided by the total time (that would be the average velocity, with an appropriate direction). If you drive one way at 60kph, turn around and drive back at 30kph for twice as long, the average speed is not zero...

But distance is a scalar... displacement is the vector... speed = distance/time. whereas velocity = displacement/time.

ie: by distance, I think he means the total distance travelled... not the magnitude of the displacement...

 

[Heat]

I forgot the main problem .

 

[learningphysics]

I forgot the main problem .

Did you calculate the total distance and total displacement? Where are you getting stuck?

 

[Heat]

Did you calculate the total distance and total displacement? Where are you getting stuck?

I am getting stuck on how to get the average speed.
You mentioned that its total distance over total time.

Total distance within 3 seconds is 9m.
Total time would have to 3s.

Would that mean that average speed is 3.0 m/s?
or would it be 2m/s for the first 2 seconds, and 3m/s for the last second, add those up for a grand total of: 7m/s, but then I feel that it should be something like this 3m - 4m/3s over 2 seconds = -1/1 = -1m/s.

For average velocity, you mentioned that its displacement over time.

The time is 3 seconds.
Displacements is 7m.

7/3= 2.33m/s?

 

[learningphysics]

How are you getting 9m?

 

[Heat]

How are you getting 9m?

It would seem that since it's within 3 seconds, and at three seconds velocity is at 3m/s.

3meters per second, which means that there are 3 seconds = 3x3= 9.

 

[learningphysics]

It would seem that since it's within 3 seconds, and at three seconds velocity is at 3m/s.

3meters per second, which means that there are 3 seconds = 3x3= 9.

No, in this case distance and displacement are the same...

In the first 2s it is traveling at 2m/s... 2s * 2m/s = 4m
In the last 1s it is traveling at 3m/s... 1s * 3m/s =3m

So a total distance of 7m.

Another way to look at it:

Take the area under the vt graph for displacement, but the areas that fall below the x-axis are negative...

And for distance take all areas as positive.

 

[Heat]

No, in this case distance and displacement are the same...

In the first 2s it is traveling at 2m/s... 2s * 2m/s = 4m
In the last 1s it is traveling at 3m/s... 1s * 3m/s =3m

So a total distance of 7m.

Another way to look at it:

Take the area under the vt graph for displacement, but the areas that fall below the x-axis are negative...

And for distance take all areas as positive.

So it would really be total distance 7m/ 3 seconds.

average speed = 7m/3s = 2.33m/s.

 

[FedEx]

Homework Statement

http://img2.freeimagehosting.net/uploads/2256719644.jpg​

A ball moves in a straight line (the x-axis). The graph in the figure shows this ball's velocity as a function of time.

What are the ball's average velocity during the first 3.0s ?

What are the ball's average speed during the first 3.0s ?

Suppose that the ball moved in such a way that the graph segment after 2.0s was -3.0 m/s instead of +3.0m/s . Find the ball's and average velocity during the first 3.0s in this case.

Suppose that the ball moved in such a way that the graph segment after 2.0s was -3.0m/s instead of +3.0 m/s. Find the ball's average speed during the first 3.0s in this case.

The Attempt at a Solution

Ok, from my reading I understand that average speed is not the magnitude of average velocity.
To determine average speed you need total total distance/ total time.
I know that if the displacement is 0, then the average velocity is 0.


What I have done so far is get the slop from time 0 through time =3s. It gives me 1.
Since, having the tangent line slop to the three seconds, would be instantaneous speed, then I would not assume that speed = 0.

My other try is that I start at the point from 0s @ 2m/s through 3s @ 3m/s.
The line would have a slope of 1/3 resulting in the average speed of .66m/s. Is this right, if not can you point me in the right direction.

For the first two equations the ansers are simple. But for the remaining questions if the velocity is -3m/s after 2 seconds then it would be a discontinuos graph. I think that for the remaining questions they have not considered the motion after 2 seconds.

 

[learningphysics]

Yup, average velocity and average speed are both 2.33m/s for the first part.

 

[learningphysics]

Were you able to do the second part with the -3.0m/s?

 

[Heat]

I did this one :Suppose that the ball moved in such a way that the graph segment after 2.0s was -3.0 m/s instead of +3.0m/s . Find the ball's and average velocity during the first 3.0s in this case.

As you mentioned that it was displacement over time,

I knew that it traveled 4m during the first 2 seconds, and 3 meters during the one second.

But sing 3 is now in the negative range it would be 4m - 3m = 1m.

1m/ 3sec = .33m/s.

I must admit, that I also thought that since the first two answers where repetitive, I would think that the average speed would be the same answer .33m/s, but I am wrong.

I would get distance over time, 4 m + -3m = 1m over time of 3-0= 3. resulting in 1/3? This can't be.

May you please explain the logic to this furthur, I thank learningphysics,FedEx,berkeman for their time helping me through this problem.

 

[learningphysics]

You did the displacement and velocity right.

In this second situation, the distance and displacement will be different... to get total distance traveled always take the speed as positive for every part... So for distance it would be just like the first problem... 2*2 + 1*3 = 7m. During the first 2s he traveled 2*2 = 4m. During the last 1s, his speed is still 3m/s (we don't care about direction so the minus is ignored here)... so the distance traveled in the last 1s is just 1*3 = 3m

So average speed is just 7/3 = 2.33m/s. For distance, direction doesn't matter. For displacement it does... So in this situation distance and average speed should be just like before...

distance is just the total length you have travelled... for example you drive 20km to the store, then you go 30km somewhere else, 50km somewhere else... So total distance is 100km. You just add them up for total distance traveled (ie that's what the speedometer on the car shows... the car doesn't know which direction you're going)... displacement will be different because that depends on the directions I travelled.

Have you studied calculus? Distance would be:

\int|\vec{v}|dt

where as displacement would be:

\int\vec{v}dt

 

[Heat]

Ok, I reread this:

Take the area under the vt graph for displacement, but the areas that fall below the x-axis are negative...

And for distance take all areas as positive.

I understand that distance will be positive,
and for displacement, I would need to take to account the - displacements.

I have not studied calculus, but I am in the process of learning along with physics. From what I could understand from the distance equation you provided is that velocity will always be postive times distance and time.

and for displacement, velocity can be either positive or negative, multiplied by distance and time.

I am really grateful to all the help I have received from learningphysics, you have been helping me understand the issues I have been having with the physics problems. Thank you.

 

[learningphysics]

Ok, I reread this:

Take the area under the vt graph for displacement, but the areas that fall below the x-axis are negative...

And for distance take all areas as positive.

I understand that distance will be positive,
and for displacement, I would need to take to account the - displacements.

I have not studied calculus, but I am in the process of learning along with physics. From what I could understand from the distance equation you provided is that velocity will always be postive times distance and time.

and for displacement, velocity can be either positive or negative, multiplied by distance and time.

Yup. Everything you posted is exactly right.

I am really grateful to all the help I have received from learningphysics, you have been helping me understand the issues I have been having with the physics problems. Thank you.

No prob. You're welcome.