# Finding Velocity: A Distance vs Time Graph Case

• deaninator
In summary, the conversation discusses finding the velocity of a point on a distance vs time graph when the slope of the line is not completely straight. It is mentioned that for each section of the graph, the velocity is constant and that there may be multiple velocities on the graph. It is suggested to compensate for the change in gradient by subtracting the time at which the gradient changed from the point on the graph. This will allow for the calculation of the velocity at a specific point on the graph. The formula V = D/T can be used, with D representing the distance and T representing the time.
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.

V = D/T

## The Attempt at a Solution

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

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

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.

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

jarednjames said:
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.

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

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

deaninator said:
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

jarednjames said:
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?

If Y = D and X = T then yes.

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

Jared

jarednjames said:
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.

## 1. What is a distance vs. time graph?

A distance vs. time graph is a visual representation of an object's motion over a period of time. It plots the distance an object travels on the y-axis and the time it takes to travel that distance on the x-axis.

## 2. How do you find velocity from a distance vs. time graph?

To find velocity from a distance vs. time graph, you need to calculate the slope of the line. This can be done by taking the change in distance (y-axis) and dividing it by the change in time (x-axis). The resulting value will give you the object's velocity at that specific point in time.

## 3. What does a steep slope on a distance vs. time graph indicate?

A steep slope on a distance vs. time graph indicates a greater velocity. This means that the object is covering more distance in a shorter amount of time, indicating a faster speed.

## 4. How can you determine if an object is at rest on a distance vs. time graph?

If an object is at rest, the distance vs. time graph will show a horizontal line. This means that the object is not moving and the slope of the line is 0. This is because the distance is not changing over time.

## 5. What other factors can affect the shape of a distance vs. time graph?

Other factors that can affect the shape of a distance vs. time graph include changes in speed and direction. If an object is speeding up or slowing down, the slope of the line will change accordingly. If the object changes direction, the slope will also change and the line may become curved.

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