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Size:  354.1 KB hi, for part a of this question, I am confused about how it's drawn. This is my working but it's really different to the right answer. Thanks.
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    Start by drawing a displacement/time graph, then consider the speed (scalar i.e. no negative values) of the ball along it's path.

    edit: saw the second pic.

    Not just at the extremes. The straight-line graph you have there is incorrect.

    And the answer to the question in the picture is that it's asking for a speed / time graph, not a velocity / time graph.
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    Yep, it's gonna be in an upside down V because it accelerates towards the ground after it's reached its peak.
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    (Original post by Alexion)
    Start by drawing a displacement/time graph, then consider the speed (scalar i.e. no negative values) of the ball along it's path.

    edit: saw the second pic.

    Not just at the extremes. The straight-line graph you have there is incorrect.

    And the answer to the question in the picture is that it's asking for a speed / time graph, not a velocity / time graph.
    Oh yh, only just realised that. This is not one of the common questions I have tried so now I am confused. I understand that velocity can be negative and speed can only be positive. But why does it have a v shape though? Also if I was asked to draw the velocity time time, would my diagram be right? Thanks.
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    (Original post by Supersaps)
    Yep, it's gonna be in an upside down V because it accelerates towards the ground after it's reached its peak.
    Hi, thanks for helping. I just checked the answer the v shape isn't upside down. Its just a v shape on the graph. I don't really understand how that works.
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    (Original post by coconut64)
    Oh yh, only just realised that. This is not one of the common questions I have tried so now I am confused. I understand that velocity can be negative and speed can only be positive. But why does it have a v shape though? Also if I was asked to draw the velocity time time, would my diagram be right? Thanks.
    Yes your diagram would be correct if that's what was asked. Remember speed is the modulus of velocity and when you draw a mod(y) graph, anything below the x axis is reflected in the x axis, so you simply take the positive value

    So take your velocity time graph you drew and reflect anything below the x axis IN the x axis. What kind of graph do you get?
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    (Original post by Supersaps)
    Yep, it's gonna be in an upside down V because it accelerates towards the ground after it's reached its peak.
    If it accelerated toward the ground after reaching the peak (which it does), the speed would increase and the line would be going upwards, so it can't be an upside down V. It would be an upright V
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    Thanks for the reply. Yeah, sorry the V is normal shaped. That was mis-worded by me otherwise we'd be talking about negative speed, which doesn't exist!


    Let's start from the 3 known parts of the graph we can get from common sense.



    - We know that speed is high at the start because it's fired upwards (This is the why the line is high at the start)

    - We know that the speed is 0 when the ball reaches the top of its flight (because all the initial speed has now been "converted" into height and there's no more speed left, that's why the line touches 0 in the middle of the V)

    - We know that the ball from its peak height with a speed of 0 will now accelerate towards the earth leading to the line being high as it hits the ground.



    Then you can basically just join those three known points together to get your graph.

    How do you know they're connected with straight lines? Because the gradient of a speed/time graph is the acceleration which is the constant 9.81m/s and a constant gradient = a straight line.

    SS
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    Moved to maths.
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    (Original post by Supersaps)
    Thanks for the reply. Yeah, sorry the V is normal shaped. That was mis-worded by me otherwise we'd be talking about negative speed, which doesn't exist!


    Let's start from the 3 known parts of the graph we can get from common sense.



    - We know that speed is high at the start because it's fired upwards (This is the why the line is high at the start)

    - We know that the speed is 0 when the ball reaches the top of its flight (because all the initial speed has now been "converted" into height and there's no more speed left, that's why the line touches 0 in the middle of the V)

    - We know that the ball from its peak height with a speed of 0 will now accelerate towards the earth leading to the line being high as it hits the ground.



    Then you can basically just join those three known points together to get your graph.

    How do you know they're connected with straight lines? Because the gradient of a speed/time graph is the acceleration which is the constant 9.81m/s and a constant gradient = a straight line.

    SS
    Oh okay. This does make sense. So the ball accelerates after it has reached its max height. But how do you know that the speed after the acceleration will be greater than 14? (This is when she catches it ). Thanks
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    (Original post by Student403)
    Yes your diagram would be correct if that's what was asked. Remember speed is the modulus of velocity and when you draw a mod(y) graph, anything below the x axis is reflected in the x axis, so you simply take the positive value

    So take your velocity time graph you drew and reflect anything below the x axis IN the x axis. What kind of graph do you get?
    So velocty time graph involves the negative part of the graph but speed time time can be only v shapes ? Also for this question, how do u knownthat the spped after acceleration will be higher than 14 (at the start) thanks.
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    (Original post by coconut64)
    Oh okay. This does make sense. So the ball accelerates after it has reached its max height. But how do you know that the speed after the acceleration will be greater than 14? (This is when she catches it ). Thanks
    Yep! The ball accelerates back down to the earth again.


    To find out the final speed, which equation do you think we have the information for? N.B. I think that the answer we get in question b is a good start.

    SS
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    (Original post by Supersaps)
    Yep! The ball accelerates back down to the earth again.


    To find out the final speed, which equation do you think we have the information for? N.B. I think that the answer we get in question b is a good start.

    SS
    Thats a gd idea but this part is the first part of the question. So they want you to draw it before doing the question. The markscheme does not specify the final speed on the graph. So how can you assume whether the final speed is lower or greater than the inital speed ? Thanks for the speedy reply by the way.
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    (Original post by coconut64)
    Thats a gd idea but this part is the first part of the question. So they want you to draw it before doing the question. The markscheme does not specify the final speed on the graph. So how can you assume whether the final speed is lower or greater than the inital speed ? Thanks for the speedy reply by the way.
    Oh, I see what you mean. Yeah, you have no idea on the final speed of the ball so pretty much finishing the V at any speed is fine. They won't mark you down for ending the line a little early or late as you're not going to know the final speed until you've done the last question.

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    (Original post by coconut64)
    Thats a gd idea but this part is the first part of the question. So they want you to draw it before doing the question. The markscheme does not specify the final speed on the graph. So how can you assume whether the final speed is lower or greater than the inital speed ? Thanks for the speedy reply by the way.
    Ball is fired from a height of 2.5 metres, i.e: it starts from 2.5 metres.

    Ball starts falling, picks up speed, it is at the same speed as it was intially launched at when it is 2.5 metres above ground, but it has another 2.5 metres to travel before it hits the ground so it picks up more speed in that time.
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    (Original post by Supersaps)
    Oh, I see what you mean. Yeah, you have no idea on the final speed of the ball so pretty much finishing the V at any speed is fine. They won't mark you down for ending the line a little early or late as you're not going to know the final speed until you've done the last question.

    SS
    Are you sure?
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    (Original post by Zacken)
    Are you sure?
    Yes.
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    (Original post by Supersaps)
    Yes.
    Air resistance is negligible... the ball falls more than it rises...
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    Oh duuuuuh.
 
 
 
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