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[Mechanics 1] Velocity-Time Differentiation

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    5 (iv) is what I don't get. I'm probably missing something obvious.

    I differentiated ds/dt and got
    v = -2t + 20
    Then put in 4 for v so
    4 = -2t + 20
    2t = 16
    t = 8
    But the answer must be > 10, and the mark scheme uses -4 instead of 4 for v. Why does it use -4?

    Thanks
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    (Original post by DarkEnergy)


    5 (iv) is what I don't get. I'm probably missing something obvious.

    I differentiated ds/dt and got
    v = -2t + 20
    Then put in 4 for v so
    4 = -2t + 20
    2t = 16
    t = 8
    But the answer must be > 10, and the mark scheme uses -4 instead of 4 for v. Why does it use -4?

    Thanks
    Because the speed is the magnitude of the velocity, and velocity is what you get from differentiation. Therefore v=\pm 4 .
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    (Original post by RDKGames)
    Because the speed is the magnitude of the velocity, and velocity is what you get from differentiation. Therefore v=\pm 4 .
    Ah yes cheers, but how do you know to use -4 or is it just trying both of them and seeing which one gives a result > 10?
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    (Original post by DarkEnergy)
    Ah yes cheers, but how do you know to use -4 or is it just trying both of them and seeing which one gives a result > 10?
    You don't. Just plug both in and see what you get. Obviously if t<10 then you disregard that answer as our domain is strictly t>10 as stated by the information we're given.
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    (Original post by RDKGames)
    You don't. Just plug both in and see what you get. Obviously if t<10 then you disregard that answer as our domain is strictly t>10 as stated by the information we're given.
    Thanks, thought there was something more complex behind it!
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    (Original post by DarkEnergy)
    Thanks, thought there was something more complex behind it!
    Nope. Though if you really don't want to try both values, you could plug in t=10 into the differential and see that the value you get is v=0. Therefore for t>10 you will get v<0 just by inspection of the equation, as the minus grows bigger.
 
 
 
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