Finding Velocity: A Distance vs Time Graph Case

[deaninator]

Homework Statement

How can I find the velocity of a point, for example (36,24) on a distance vs time graph when the slope of the line is not completely straight? On the graph, the slope starts from zero shoots upwards in one direction, then breaks the path and shoots in a slightly different direction. The line does not form a curve.

Homework Equations

V = D/T

The Attempt at a Solution

V = 36/24...but that is incorrect.

 

[JaredJames]

Speed = Distance / Time

If the lines are straight (even in sections), for each section the velocity is constant.

There are two (or more) different velocities on this graph. Each section (each gradient) has its own velocity.

For the first section, V = D/T will work, however, for the second section (once the line gradient has changed) you have to compensate.

So your T value = the point on the graph minus the time when the graph changed gradient. And the same for your distance graph.

So, if your first line gradient is constant until T = 10s, then your time value for the second velocity = point on graph (eg 15s) minus the point at which the gradient changed (10s). Which means you get 15 - 10 = 5 seconds.

Sorry it isn't too clear, let me know if you need more.

Jared

 

[fss]

Just add the areas under the various sections of your velocity graph. Just because you can't determine the area in one fell swoop doesn't mean you are unable to do the problem.

 

[JaredJames]

He doesn't want the whole velocity, but that at a specific point.

 

[deaninator]

Speed = Distance / Time

If the lines are straight (even in sections), for each section the velocity is constant.

There are two (or more) different velocities on this graph. Each section (each gradient) has its own velocity.

For the first section, V = D/T will work, however, for the second section (once the line gradient has changed) you have to compensate.

So your T value = the point on the graph minus the time when the graph changed gradient. And the same for your distance graph.

So, if your first line gradient is constant until T = 10s, then your time value for the second velocity = point on graph (eg 15s) minus the point at which the gradient changed (10s). Which means you get 15 - 10 = 5 seconds.

Sorry it isn't too clear, let me know if you need more.

Jared

Um ok. So what happens if the slope of the line is straight until (10,5). The at point (10,5), the line shoots in a different different direction and your asked to find point A which is (20, 10)? Find the velocity of A that is.

 

[deaninator]

He doesn't want the whole velocity, but that at a specific point.

Yes he just wants the velocity of that specific point (A).

 

[JaredJames]

Um ok. So what happens if the slope of the line is straight until (10,5). The at point (10,5), the line shoots in a different different direction and your asked to find point A which is (20, 10)? Find the velocity of A that is.

Well you know V = D / T

So for this point (20 , 10) you would need to remove everything up to (10 , 5):

For the X coordinate = (20 - 10)

For the Y coordinate = (10 - 5)

This way you remove the first velocity and are only left with the second section of the graph.

The you have X and Y (which correspond to your D and T whichever way you have them).

Jared

 

[deaninator]

Well you know V = D / T

So for this point (20 , 10) you would need to remove everything up to (10 , 5):

For the X coordinate = (20 - 10)

For the Y coordinate = (10 - 5)

This way you remove the first velocity and are only left with the second section of the graph.

The you have X and Y (which correspond to your D and T whichever way you have them).

Jared

So basically then 5/10 to find the velocity of A?

 

[JaredJames]

If Y = D and X = T then yes.

As long as you understand how you come to that answer.

Jared

 

[deaninator]

If Y = D and X = T then yes.

As long as you understand how you come to that answer.

Jared

Thanks, I need to look into it more then I'll get back to you.