Help in drawing a displacement time graph..

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TSR George
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#1
Report Thread starter 7 years ago
#1
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?

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
Report 7 years ago
#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.
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TSR George
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#3
Report Thread starter 7 years ago
#3
(Original post by 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.
Sorry for not clearing that up, I was only drawing the displacement graph for his downwards vertical displacement. He travels -1.25 metres in half a second, and because G is approximated to 10 m/s-2 his final velocity should be - 5.

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
Report 7 years ago
#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 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
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#5
(Original post by 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.
Firstly let me thank you for your help, I really appreciate it.

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

Image
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Stonebridge
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#6
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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.
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TSR George
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#7
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#7
(Original post by 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.
Yes I get that the objects actual velocity will be the vector sum of its vert. and horizontal velocities.

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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Stonebridge
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#8
Report 7 years ago
#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.
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TSR George
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#9
Report Thread starter 7 years ago
#9
(Original post by 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.
Could you go a bit further as to why it is curved? I know how to find the gradient of a curved graph and yes I understand that his actual velocity will be different because we would have to add his horizontal speed.
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Stonebridge
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#10
Report 7 years ago
#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.

Can you post the original question please.
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TSR George
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#11
Report Thread starter 7 years ago
#11
(Original post by 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.
I am trying to find it! We were supposed to draw it even after that. Id just like to know why it must not be straight, and instead smooth/curved?
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Stonebridge
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#12
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#12
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.
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TSR George
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#13
Report Thread starter 7 years ago
#13
(Original post by Stonebridge)
On a curve the gradient of the graph will be changing to show that the velocity (equal to gradient) is changing.
Alright I finally got it, because it is accelerating the speed component is increasing, therefore the time it takes to travel a distance is decreasing in proportion so the vertical component of the displacement is increasing more relative to the time, so it slopes down. Stupid, corny but I think I got it. Will rep you thank you so much.
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