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    Why do you have to find a new value of t (in part b)? Why can't you just use the t value found for the x equation in part a, as we use the same p value for both equations (-2/5) so why isn't this the same for the value of t?

    Thanks
    (hope the photo isn't too hard to see lol)

    *Edit: My laptop won't let me upload a photo, so here's the link, if that's okay: http://pmt.physicsandmathstutor.com/...hapter%202.pdf
    and it's exercise 2B question 5

    Thanks again!
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    (Original post by sportyegg)
    Why do you have to find a new value of t (in part b)? Why can't you just use the t value found for the x equation in part a, as we use the same p value for both equations (-2/5) so why isn't this the same for the value of t?

    Thanks
    (hope the photo isn't too hard to see lol)

    *Edit: My laptop won't let me upload a photo, so here's the link, if that's okay: http://pmt.physicsandmathstutor.com/...hapter%202.pdf
    and it's exercise 2B question 5

    Thanks again!
    Because t is the independent variable here, and the x,y are dependent variables. The t value at the x-intercept will not be the same as the one at the y-intercept.

    We reuse p because that’s a fixed constant which we determine in part a
 
 
 
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