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    A guide to practical work...its a graph question
    In the First picture you can see the questions.I cant understand the second question
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    In the second picture is the Answers and you can see answer 2
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    how to find the equation of the line and
    why the gradient is 2/a

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    (Original post by reb0xx)
    A guide to practical work...its a graph question
    In the First picture you can see the questions.I cant understand the second question
    In the second picture is the Answers and you can see answer 2
    how to find the equation of the line and
    why the gradient is 2/a

    On the Y-axis you have t^2, on the X-axis - s. So you can put y\equiv t^2 and x\equiv s to obtain the equation of the line:

    y=\alpha x+\beta.

    You also expect that t^2=\dfrac{2}{a}s\; \Longleftrightarrow \; y=\dfrac{2}{a}x.

    From this you can see that the gradient \alpha is 2/a

    To find the equation of the line, you would have to

    (1) find the gradient \alpha;

    (2) find \beta. To find \beta, put coordinates of a point on the line into the equation of the line.
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    (Original post by jaroc)
    On the Y-axis you have t^2, on the X-axis - s. So you can put y\equiv t^2 and x\equiv s to obtain the equation of the line:

    y=\alpha x+\beta.

    You also expect that t^2=\dfrac{2}{a}s\; \Longleftrightarrow \; y=\dfrac{2}{a}x.

    From this you can see that the gradient \alpha is 2/a

    To find the equation of the line, you would have to

    (1) find the gradient \alpha;

    (2) find \beta. To find \beta, put coordinates of a point on the line into the equation of the line.
    thank you!!!
    but wouldn't be easier if we solve the equation?
 
 
 
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