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    Or was it from an acceleration-time graph?? Please tell me how to find it from both types of graph if it's possible. Thank you!
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    (Original post by Ryhana)
    Or was it from an acceleration-time graph?? Please tell me how to find it from both types of graph if it's possible. Thank you!
    speed = distance / time, which can be written as dx/dt.

    acceleration = rate of change of velocity = dv/dt. Gotta be slightly careful when it comes to speed vs velocity.

    Everything follows from there (the relationship between speed and acceleration, how to get to different ones by integrating/differentiating etc).
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    (Original post by Ryhana)
    Or was it from an acceleration-time graph?? Please tell me how to find it from both types of graph if it's possible. Thank you!
    From a velocity-time graph: note that where v is velocity and x is displacement, v = dx/dt.
    Thus x = integral of v dt, so the numerical value of the displacement will be the area under the graph, since integrals represent area under a curve.

    It isn't directly possible to find displacement from an acceleration-time graph, unless you are given or can find the equation of the graph.
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    (Original post by SeanFM)
    speed = distance / time, which can be written as dx/dt.

    acceleration = rate of change of velocity = dv/dt. Gotta be slightly careful when it comes to speed vs velocity.

    Everything follows from there (the relationship between speed and acceleration, how to get to different ones by integrating/differentiating etc).
    (Original post by HapaxOromenon3)
    From a velocity-time graph: note that where v is velocity and x is displacement, v = dx/dt.
    Thus x = integral of v dt, so the numerical value of the displacement will be the area under the graph, since integrals represent area under a curve.

    It isn't directly possible to find displacement from an acceleration-time graph, unless you are given or can find the equation of the graph.
    Is there a method which doesn't involve integration/differentiation?
    I do AQA maths and we don't use calculus in M1. I only found this type of question (to find the displacement) from other exam board past papers. If there is no alternative besides integration/differentiation, I suppose this won't come up on my AQA M1 exam?
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    (Original post by Ryhana)
    Is there a method which doesn't involve integration/differentiation?
    I do AQA maths and we don't use calculus in M1. I only found this type of question to find the displacement from other exam board past papers. If there is no alternative besides integration/differentiation, I suppose this won't come up on my AQA M1 exam?
    The specification states knowledge from C1 and C2, which includes calculus, is assumed.
    If the graph is made up of straight lines then you can find the area under it using formulae for area of a trapezium, rectangle, triangle, etc. from GCSE.
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    (Original post by Ryhana)
    Is there a method which doesn't involve integration/differentiation?
    I do AQA maths and we don't use calculus in M1. I only found this type of question (to find the displacement) from other exam board past papers. If there is no alternative besides integration/differentiation, I suppose this won't come up on my AQA M1 exam?
    You could find the area (which is what integration does) in other ways, like finding shapes and their areas.


    Eg if you had a velocity-time graph. starting at 0 and had a gradient of 1 (i.e an acceleration of 1 m/s^2) and were asked to find the displacement after 10 seconds, then you could either use SUVAT or find the area between t=0 and t=10 which is just a triangle, so area = 1/2 * base * height = 1/2 * 10 * 10 = 50m.
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    (Original post by HapaxOromenon3)
    The specification states knowledge from C1 and C2, which includes calculus, is assumed.
    If the graph is made up of straight lines then you can find the area under it using formulae for area of a trapezium, rectangle, triangle, etc. from GCSE.
    (Original post by SeanFM)
    You could find the area (which is what integration does) in other ways, like finding shapes and their areas.


    Eg if you had a velocity-time graph. starting at 0 and had a gradient of 1 (i.e an acceleration of 1 m/s^2) and were asked to find the displacement after 10 seconds, then you could either use SUVAT or find the area between t=0 and t=10 which is just a triangle, so area = 1/2 * base * height = 1/2 * 10 * 10 = 50m.
    Sorry, I think my problem is not so much finding the area but it is when the velocity-time graph goes into the negative axis.
    So if the question asks to find the displacement and there are two areas (above and below) the x-axis, are we supposed to subtract the two areas to find the displacement?

    And I'm also confused between when line has a negative gradient and when the line crosses the x-axis. Which one means that the 'object' is moving in the other direction?
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    (Original post by Ryhana)
    Sorry, I think my problem is not so much finding the area but it is when the velocity-time graph goes into the negative axis.
    So if the question asks to find the displacement and there are two areas (above and below) the x-axis, are we supposed to subtract the two areas to find the displacement?

    And I'm also confused between when line has a negative gradient and when the line crosses the x-axis. Which one means that the 'object' is moving in the other direction?
    In response to the first question: Yes.
    In response to the second question: It is going in the other direction when the line is below the x-axis.
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    (Original post by HapaxOromenon3)
    In response to the first question: Yes.
    In response to the second question: It is going in the other direction when the line is below the x-axis.
    Thank you!
 
 
 
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