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Math Help - a body is coasting to a stop and the only force acting on it is a resistance

  1. #1
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    a body is coasting to a stop and the only force acting on it is a resistance

    A body is coasting to a stop and the only force acting on it is a resistance proportional to its speed, according to the equation:
    \frac{ds}{dt}=v_f=v_0e^{(\frac{-kt}{m})} ; s(0)=0, where v_0 is the body's initial velocity (in m/s), v_f is its final velocity, m is its mass, k is a constant, and t is time.

    (a) If a body with mass m = 50 kg and k = 1.5 kg/sec initially has a velocity of 30 m/s, how long, to the nearest second, will it take to slow to 1 m/s?

    (b) How far, to the nearest 10 meters, will the body coast during the time takes to slow from 30 m/s to 1 m/s?

    (c) If the body coasts from 30 m/s to a stop, how far will it coast?
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    You can get (a) by using the given differential equation and solving for t.

    Since v_0\text{, }k,\text{ and }m are constants, you have \frac{ds}{dt} equal to a function of t, so you only have to take the integral of both sides to get s as a function of t.

    If you don't know how to integrate, try setting s=c_1e^{c_2t}+C, take the derivative and choose values of c_1\text{ and }c_2 to make the equation work. You'll have to use the initial condition s(0)=0 to get C.

    The answer to (c) is that the question is invalid - if you look at the differential equation, you see that v_f never reaches zero, so it never stops. But you can take the limit as t goes to infinity of the equation you found in (b) to get the result.

    Post again in this thread if you're still having trouble.
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  3. #3
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    Quote Originally Posted by yoman360 View Post
    A body is coasting to a stop and the only force acting on it is a resistance proportional to its speed, according to the equation:
    \frac{ds}{dt}=v_f=v_0e^{(\frac{-kt}{m})} ; s(0)=0, where v_0 is the body's initial velocity (in m/s), v_f is its final velocity, m is its mass, k is a constant, and t is time.

    (a) If a body with mass m = 50 kg and k = 1.5 kg/sec initially has a velocity of 30 m/s, how long, to the nearest second, will it take to slow to 1 m/s?

    (b) How far, to the nearest 10 meters, will the body coast during the time takes to slow from 30 m/s to 1 m/s?

    (c) If the body coasts from 30 m/s to a stop, how far will it coast?

    (a) solve the equation v(t) = v_0 e^{-\frac{kt}{m}} for t , then sub in your given values to calculate t


    (b) using t found in part (a) as the upper limit of integration ...

    d = \int_0^t v(t) \, dt


    (c) d = \lim_{b \to \infty} \int_0^b v(t) \, dt
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    Quote Originally Posted by skeeter View Post
    (a) solve the equation v(t) = v_0 e^{-\frac{kt}{m}} for t , then sub in your given values to calculate t


    (b) using t found in part (a) as the upper limit of integration ...

    d = \int_0^t v(t) \, dt


    (c) d = \lim_{b \to \infty} \int_0^b v(t) \, dt
    I got t=0 but i think i did something wrong
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    I assume you mean for (a). You should be setting v_0=30\text{, }k=1.5\text{, }m=50\text{, and }v_f=1. If you still get t=0, please show us your work and one of us will show you where you went wrong.

    - Hollywood
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    Quote Originally Posted by hollywood View Post
    I assume you mean for (a). You should be setting v_0=30\text{, }k=1.5\text{, }m=50\text{, and }v_f=1. If you still get t=0, please show us your work and one of us will show you where you went wrong.

    - Hollywood
    v_f=v_0e^{(\frac{-kt}{m})}

    1=30e^{(\frac{-1.5t}{50})}

    when making this post i realized the mistake i made

    now i get

    t=\frac{-50}{1.5}*ln(\frac{1}{30})

    (A) t≈ 133 seconds

    hopefully thats correct
    Last edited by yoman360; March 27th 2010 at 10:13 AM.
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  7. #7
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    Quote Originally Posted by yoman360 View Post
    v_f=v_0e^{(\frac{-kt}{m})}

    1=30e^{(\frac{-1.5t}{50})}

    when making this post i realized the mistake i made

    now i get

    t=\frac{-50}{1.5}*ln(\frac{1}{30})

    hopefully thats correct
    correct ... that is the value of t such that v(t) = 1
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