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    Can someone explain to me why this is wrong?
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    Going to have to elaborate
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    The second one is a velocity time graph of a ball decelerating as it goes up then acellerating in the opposite direction (makes it negative)as it goes down also when graph is 0 ball has 0 velocity
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    (Original post by BDunlop)
    Going to have to elaborate
    Sorry about that I thought I'd edited the question.. doesn't look like it worked.

    Well the v/t graph is the 'original' and then we had to draw the s/t and a/t graphs for it. But apparently the s/t and a/t graphs I drawn are wrong and I'm having trouble understanding why.
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    (Original post by Hasham123)
    The second one is a velocity time graph of a ball decelerating as it goes up then acellerating in the opposite direction (makes it negative)as it goes down also when graph is 0 ball has 0 velocity
    That's what I thought it was representing but apparently it's wrong. I was told that once the line went below the x-axis the object would still be decelerating because of the negative gradient.
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    I'm slightly confused aswell, where's the original v/t graph?
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    (Original post by Kamila_Kamizaki)
    Can someone explain to me why this is wrong?
    Remember that the gradient of v/t is the acceleration. Your a/t graph looks right. The s/t graph, it starts at a maximum speed and slows down at a constant rate, then reaches s=0 then speeds up again at the same rate it decelerated (straight line) so it will look like a V
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    (Original post by BDunlop)
    Okay I see, well you've draw for your s/t graph a negative speed, but speed is a scalar so can't have direction can it, presuming that was your second one?
    For the acceleration, remeber it's the gradient of the velocity/time graph. Now, looking at the v/t graph it's obvious the gradient changes so the acceleration time graph can't be a horizontal line can it. The a/t graph starts with maximum acceleration in the negative direction then becomes 0 when it reaches maximum displacement. I would say the middle graph if the line started in the negative sense would be okay for acceleration. Remember gravity is constant so the rate of change of acceleration should be zero..
    Sorry, I should've made this clear by s/t graph I meant displacement. Also how does the gradient change in the second (v/t) graph? It's a straight line so the gradient is constant isn't it? Also how does gravity come in to the question? There are no details about the object that is moving and what it is.
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    (Original post by BDunlop)
    I'm slightly confused aswell, where's the original v/t graph?
    Ah, from top to bottom: displacement/time, velocity/time, acceleration/time.
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    (Original post by Kamila_Kamizaki)
    Sorry, I should've made this clear by s/t graph I meant displacement. Also how does the gradient change in the second (v/t) graph? It's a straight line so the gradient is constant isn't it? Also how does gravity come in to the question? There are no details about the object that is moving and what it is.
    Just ignore that/delete it look up
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    For the displacement/time it's displacement will start increasing into the positive sense like you drew, then it will reach max like you said, then decrease again to s=0 from where it was thrown. But you're right to draw a curved line because gradient for s/t is velocity. Looks fine
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    (Original post by BDunlop)
    For the displacement/time it's displacement will start increasing into the positive sense like you drew, then it will reach max like you said, then decrease again to s=0 from where it was thrown. But you're right to draw a curved line because gradient for s/t is velocity. Looks fine
    Okay thank you for your help - and sorry for the misunderstandings!!
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    (Original post by Kamila_Kamizaki)
    That's what I thought it was representing but apparently it's wrong. I was told that once the line went below the x-axis the object would still be decelerating because of the negative gradient.
    Negative acceleration is decelleration
 
 
 
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