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    (Original post by Jozanic)
    Still getting the same equation
    You might want to practice some algebraic manipulation soon then. Anyhow:

    Gonna use t_2=60-\frac{2}{3}t_2

    1000=\frac{1}{2}Vt_1+V(t_2-t_1)+\frac{1}{2}V(60-t_2)

    \Rightarrow 1000=\frac{1}{2}V(\frac{V}{3})+V  (60-\frac{2}{3}V-\frac{V}{3})+\frac{1}{2}V(60-60+\frac{2}{3}V)

    \Rightarrow 1000=\frac{V^2}{6}+V(60-V)+\frac{1}{3}V^2

    \Rightarrow 1000=\frac{V^2}{6}-V^2+60V+\frac{1}{3}V^2

    can you finish from there?
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    (Original post by RDKGames)
    You might want to practice some algebraic manipulation soon then. Anyhow:

    Gonna use t_2=60-\frac{2}{3}t_2

    1000=\frac{1}{2}Vt_1+V(t_2-t_1)+\frac{1}{2}V(60-t_2)

    \Rightarrow 1000=\frac{1}{2}V(\frac{V}{3})+V  (60-\frac{2}{3}V-\frac{V}{3})+\frac{1}{2}V(60-60+\frac{2}{3}V)

    \Rightarrow 1000=\frac{V^2}{6}+V(60-V)+\frac{1}{3}V^2

    \Rightarrow 1000=\frac{V^2}{6}-V^2+60V+\frac{1}{3}V^2

    can you finish from there?
    Thanks man. I made a silly mistake with one fraction. I got 20 and 100 for V.
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    Are you doing the MEI exam board, because I remember an extremely similar question from when I went through the textbook.
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    (Original post by Jozanic)
    Thanks man. I made a silly mistake with one fraction. I got 20 and 100 for V.
    That is correct, but only one of them is valid.

    Remember that there is T for which the speed is constant, and this lies between t_1 and t_2

    So T=t_2-t_1 and express it in terms of V. Try both values of V and you will see that one of them does not make sense, hence you get left with one.
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    (Original post by Jozanic)
    Thanks man. I made a silly mistake with one fraction. I got 20 and 100 for V.
    Omg that's what I got whoop whopp
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    Honestly think I'd personally learn better from just looking at a full solution...
    But I guess the forum rules do make sense, as people could use it as a cheating mechanism, however I'd work better with full solutions just saying maybe the rule could be refined or something, like to get a full solution answer, you must show a reasonable attempt, and then only can people reply with full solutions
 
 
 
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