# Help in drawing a displacement time graph..

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**TSR George**

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Hi I had a question where a biker rides off a higher point and ends up travelling 10 metres horizontally and 1.25 metres downwards.

Now it asked what his initial speed was when at the high point. Obviously this was a projectile motion vector scenario? with his horizontal speed unknown but his downwards vertical acceleration being g. He had no vertical speed at first.

I got his initial speed as 20 in the horizontal direction and the time he took to land at the 10 metres being half a second, and when he hits the ground his speed is 5 metres/second downwards. Which is correct.

When I draw the displacement time graph for his journey, it goes into the negative 1.25 metres with the time being 0.5 seconds, taking the origin to be the high point. This is also correct, as at 0.5 seconds he lands.

From what I know his instantaneous velocity is 5 ms1 at half a second..

The problem is when I checked the gradient of the graph I am getting 2.5 m/s-1 which isnt his final velocity, his final is 5 m/s-1 (taking g as 10, starting from 0)..I dont understand? isnt the velocity the gradient of a displacement time graph? Have I done something wrong?

Now it asked what his initial speed was when at the high point. Obviously this was a projectile motion vector scenario? with his horizontal speed unknown but his downwards vertical acceleration being g. He had no vertical speed at first.

I got his initial speed as 20 in the horizontal direction and the time he took to land at the 10 metres being half a second, and when he hits the ground his speed is 5 metres/second downwards. Which is correct.

When I draw the displacement time graph for his journey, it goes into the negative 1.25 metres with the time being 0.5 seconds, taking the origin to be the high point. This is also correct, as at 0.5 seconds he lands.

From what I know his instantaneous velocity is 5 ms1 at half a second..

The problem is when I checked the gradient of the graph I am getting 2.5 m/s-1 which isnt his final velocity, his final is 5 m/s-1 (taking g as 10, starting from 0)..I dont understand? isnt the velocity the gradient of a displacement time graph? Have I done something wrong?

**Edit: does the gradient of a displacement time graph only givs you the average velocity...still a bit confused. It should be 5 ms-1 at half a second but wth**
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Stonebridge

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#2

Without seeing your graph and what you did it's impossible to know what mistake you made.

His final velocity is the vector sum of his horizontal velocity (constant at 10m/s) and his downwards vertical velocity under gravity.

It seems like (but I can't be sure) that you are talking about just the vertical displacement time graph. This will only give the vertical velocity component. You still have the 10m/s horizontal to add.

As I say, it's difficult to know without seeing what you did, exactly.

His final velocity is the vector sum of his horizontal velocity (constant at 10m/s) and his downwards vertical velocity under gravity.

It seems like (but I can't be sure) that you are talking about just the vertical displacement time graph. This will only give the vertical velocity component. You still have the 10m/s horizontal to add.

As I say, it's difficult to know without seeing what you did, exactly.

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**TSR George**

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#3

(Original post by

Without seeing your graph and what you did it's impossible to know what mistake you made.

His final velocity is the vector sum of his horizontal velocity (constant at 10m/s) and his downwards vertical velocity under gravity.

It seems like (but I can't be sure) that you are talking about just the vertical displacement time graph.

As I say, it's difficult to know without seeing what you did, exactly.

**Stonebridge**)Without seeing your graph and what you did it's impossible to know what mistake you made.

His final velocity is the vector sum of his horizontal velocity (constant at 10m/s) and his downwards vertical velocity under gravity.

It seems like (but I can't be sure) that you are talking about just the vertical displacement time graph.

As I say, it's difficult to know without seeing what you did, exactly.

But on the graph, there are only two points, his negative displacement is -1.25, and the time being half a second. Since 1.25 divided by half is giving us the gradient, which is 2.5 m s-1. But I am saying that I know for a fact that his downward vertical velocity is 5 m/s-1 at half a second under the force of G, because he started from 0 speed in the vertical! so shoudnt the gradient be 5? If we double 2.5 we get that but I dont think we are supposed to double anything after we've found the gradient.

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Stonebridge

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#4

You are not measuring the gradient correctly as far as I can see. (Without your graph it's difficult to answer.)

If you join the start and end points with a straight line that is not the gradient of the graph at the point where he lands. The complete graph is a curve, for a start, not a straight line. The gradient you need to draw is the

A diagram would be useful.

If you join the start and end points with a straight line that is not the gradient of the graph at the point where he lands. The complete graph is a curve, for a start, not a straight line. The gradient you need to draw is the

**tangent**to the curve at the point where he lands.A diagram would be useful.

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**TSR George**

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#5

(Original post by

You are not measuring the gradient correctly as far as I can see. (Without your graph it's difficult to answer.)

If you join the start and end points with a straight line that is not the gradient of the graph at the point where he lands. The complete graph is a curve, for a start, not a straight line. The gradient you need to draw is the

A diagram would be useful.

**Stonebridge**)You are not measuring the gradient correctly as far as I can see. (Without your graph it's difficult to answer.)

If you join the start and end points with a straight line that is not the gradient of the graph at the point where he lands. The complete graph is a curve, for a start, not a straight line. The gradient you need to draw is the

**tangent**to the curve at the point where he lands.A diagram would be useful.

Secondly heres the graph lol. So this describes his vertical motion which is downwards 1.25 metres, he reaches the ground at half a second and thus his vertical speed is supposed to be 5 m s-1, because

V = u +at ; U = 0, T = 0.5, A = 10,

V = 10 * 0.5

V = 5 ms-1

but when I draw the displacement time graph and find its gradient at 0.5 seconds it is coming to 2.5 ms-1

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Stonebridge

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#6

Firstly

The displacement time graph is not a straight line. (As I have said before)

It's a curve.

Secondly, if you draw one only for the vertical part of the motion, you are not going to get a value for the actual motion, because you have to inlude the horizontal component. (As I said before)

Can you please post the complete original question.

Can you also draw the graph asked for (if it is asked for) which is a displacement time graph and is a curve.

Don't get it confused with the velocity time graph.

The actual velocity of the object is always the vector sum of its vertical velocity combined with the constant horizontal velocity component.

The displacement time graph is not a straight line. (As I have said before)

It's a curve.

Secondly, if you draw one only for the vertical part of the motion, you are not going to get a value for the actual motion, because you have to inlude the horizontal component. (As I said before)

Can you please post the complete original question.

Can you also draw the graph asked for (if it is asked for) which is a displacement time graph and is a curve.

Don't get it confused with the velocity time graph.

The actual velocity of the object is always the vector sum of its vertical velocity combined with the constant horizontal velocity component.

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**TSR George**

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#7

(Original post by

Firstly

The displacement time graph is not a straight line. (As I have said before)

It's a curve.

Secondly, if you draw one only for the vertical part of the motion, you are not going to get a value for the actual motion, because you have to inlude the horizontal component. (As I said before)

Can you please post the complete original question.

Can you also draw the graph asked for (if it is asked for) which is a displacement time graph and is a curve.

Don't get it confused with the velocity time graph.

Edit: I just want to find his vertical motion downwards.

The actual velocity of the object is always the vector sum of its vertical velocity combined with the constant horizontal velocity component.

**Stonebridge**)Firstly

The displacement time graph is not a straight line. (As I have said before)

It's a curve.

Secondly, if you draw one only for the vertical part of the motion, you are not going to get a value for the actual motion, because you have to inlude the horizontal component. (As I said before)

Can you please post the complete original question.

Can you also draw the graph asked for (if it is asked for) which is a displacement time graph and is a curve.

Don't get it confused with the velocity time graph.

Edit: I just want to find his vertical motion downwards.

The actual velocity of the object is always the vector sum of its vertical velocity combined with the constant horizontal velocity component.

So this is the graph purely for his vertical displacement mr bridge

So the displacement time graph. Can you explain why it would be curved? is it because of the acceleration? i will find and post the question later

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#8

Yes. The displacement time graph is a curve because of the acceleration.

The gradient of this graph, at a point, gives the velocity at that point.

To find the gradient at a point you have to draw a tangent to the curve at that point.

To find the velocity when it strikes the ground, you have to draw the tangent at that point.

If the graph is just for the vertical part of the motion, then the velocity you find is just the vertical component, and not the actual velocity.

The actual velocity is the vertical component of velocity at that point added to the horizontal component.

The gradient of this graph, at a point, gives the velocity at that point.

To find the gradient at a point you have to draw a tangent to the curve at that point.

To find the velocity when it strikes the ground, you have to draw the tangent at that point.

If the graph is just for the vertical part of the motion, then the velocity you find is just the vertical component, and not the actual velocity.

The actual velocity is the vertical component of velocity at that point added to the horizontal component.

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**TSR George**

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#9

(Original post by

Yes. The displacement time graph is a curve because of the acceleration.

The gradient of this graph, at a point, gives the velocity at that point.

To find the gradient at a point you have to draw a tangent to the curve at that point.

To find the velocity when it strikes the ground, you have to draw the tangent at that point.

If the graph is just for the vertical part of the motion, then the velocity you find is just the vertical component, and not the actual velocity.

The actual velocity is the vertical component of velocity at that point added to the horizontal component.

**Stonebridge**)Yes. The displacement time graph is a curve because of the acceleration.

The gradient of this graph, at a point, gives the velocity at that point.

To find the gradient at a point you have to draw a tangent to the curve at that point.

To find the velocity when it strikes the ground, you have to draw the tangent at that point.

If the graph is just for the vertical part of the motion, then the velocity you find is just the vertical component, and not the actual velocity.

The actual velocity is the vertical component of velocity at that point added to the horizontal component.

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#10

I don't think we are getting very far with this.

I don't see why, if you are given the values in your 1st post, you need to draw a graph and find a gradient to find velocity. It is found using suvat equations and adding components.

None of this makes much sense.

I don't see why, if you are given the values in your 1st post, you need to draw a graph and find a gradient to find velocity. It is found using suvat equations and adding components.

None of this makes much sense.

**Can you post the original question please.**
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**TSR George**

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#11

(Original post by

I don't think we are getting very far with this.

I don't see why, if you are given the values in your 1st post, you need to draw a graph and find a gradient to find velocity. It is found using suvat equations and adding components.

None of this makes much sense.

**Stonebridge**)I don't think we are getting very far with this.

I don't see why, if you are given the values in your 1st post, you need to draw a graph and find a gradient to find velocity. It is found using suvat equations and adding components.

None of this makes much sense.

**Can you post the original question please.**
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A straight displacement time graph shows uniform (constant) velocity.

If the object is accelerating (as we decided before) the graph will be curved.

On a curve the gradient of the graph will be changing to show that the velocity (equal to gradient) is changing.

If the object is accelerating (as we decided before) the graph will be curved.

On a curve the gradient of the graph will be changing to show that the velocity (equal to gradient) is changing.

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**TSR George**

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#13

(Original post by

On a curve the gradient of the graph will be changing to show that the velocity (equal to gradient) is changing.

**Stonebridge**)On a curve the gradient of the graph will be changing to show that the velocity (equal to gradient) is changing.

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