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Math Help - Argh!! Really stuck!! Please help

  1. #1
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    Argh!! Really stuck!! Please help

    Hi all, I'm having a few problems doing practical problems. I can solve some of them but I'm having problems with deaccelleration in particular.. Can you help me solve these problems?

    The distance s travelled by a train in t seconds after it applies its brakes is given by . How far has the train travelled before it comes to rest ?

    I'm totally stumped by this one but I can do some acceleration questions, such as:

    A missile is fired vertically upwards and its height h in metres after time t is given by . Calculate the maximum height reached by the missile.

    I know thw anwer to this is 1000.. But I'm lost with deaccelleration..

    Can someone please help, thanks in advance.
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  2. #2
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    If s = 16t-0.4t^2 then v (velocity) = \frac{ds}{dt} = 16-0.8t
    v=0 when the train is at rest. So solve for t. Then plug this value of t in to the distance equation to find the total distance traveled.


    h=500+100t-5t^2
    To get v, differentiate. So v = 100-10t
    You should know the equation for h represents the graph of a parabola. So for the maximum h the derivative, v, is 0. Solve v=0, plug the t value into the equation for h and you will have your answer of 1000m.
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  3. #3
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    I've tried to find t but I don't know what it is?

    I tried using 0.8 as t and using the formula as:

    s= 16t - 0.4t to be squared
    s= 16(0.8) - 0.4(0.8) to be squared but the answer I got was 12.8 and that's wrong...

    Could you please tell me the answer and how you found it..? Thanks.
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  4. #4
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    Quote Originally Posted by nzmathman View Post
    If s = 16t-0.4t^2 then v (velocity) = \frac{ds}{dt} = 16-0.8t

    v=0 when the train is at rest. So solve for t. Then plug this value of t in to the distance equation to find the total distance traveled.
    [snip]
    Quote Originally Posted by yummyburgers View Post
    I've tried to find t but I don't know what it is?

    I tried using 0.8 as t and using the formula as:

    s= 16t - 0.4t to be squared
    s= 16(0.8) - 0.4(0.8) to be squared but the answer I got was 12.8 and that's wrong...

    Could you please tell me the answer and how you found it..? Thanks.
    When you were instructed to solve for t, that meant solve v = 0 for t, that is, solve \frac{ds}{dt} = 0 for t, that is, solve 16 - 0.8t = 0 for t.
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  5. #5
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    So that would mean ds/dt = 16t -0.4t to be squared..?

    I tried subtracting 16 - 0.4 = 15.6 and 16 - 0.8 = 15.2but those answers are wrong?
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  6. #6
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    Quote Originally Posted by yummyburgers View Post
    So that would mean ds/dt = 16t -0.4t to be squared..?

    I tried subtracting 16 - 0.4 = 15.6 and 16 - 0.8 = 15.2but those answers are wrong?
    Quote Originally Posted by mr fantastic View Post
    When you were instructed to solve for t, that meant solve v = 0 for t, that is, solve \frac{ds}{dt} = 0 for t, that is, solve
    16 - 0.8t = 0 for t.
    Can you solve the red equation for t? Can you substitute that value of t into s = 16 t - 0.4 t^2 ?
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