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Math Help - velocity help

  1. #1
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    velocity help

    I solved the problem but I want to make sure I did it correctly.

    starting at time t=0, a train is halted by a braking deceleration of \frac{1}{2}\frac{m}{s^2}, with the result that its position at time t is s(t) = s_{0} +v_{0} -\frac{1}{4}(t^2)

    show that s_{0} is the train's postion when the brakes are applied and v_{0} is its velocity at that time. Find the stopping distances if v_{0} is 20 \frac {m}{s}

    s_{0} +v_{0}*(0) -\frac{1}{4}(0^2) = 0 gives s_{0} =0

    and

    v_{0} -\frac{1}{2}(0) = 0 gives v_{0} =0

    and solve s(t) to get the distance using v_{0} = 20


    20 -\frac{1}{2}t\Rightarrow t =40

    s(t) = 0 + 20(40) -\frac {1}{4}(40^2) = 400

    thus the answer is 400 metres.

    ps. teaching myself LaTex
    Last edited by Jonroberts74; October 11th 2013 at 05:22 PM.
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  2. #2
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    Re: velocity help

    Quote Originally Posted by Jonroberts74 View Post
    I solved the problem but I want to make sure I did it correctly.

    starting at time t=0, a train is halted by a braking deceleration of \frac{1}{2}\frac{m}{s^2}, with the result that its position at time t is s(t) = s_{0} +v_{0} -\frac{1}{4}(t^2)

    show that s_{0} is the train's postion when the brakes are applied and v_{0} is its velocity at that time. Find the stopping distances if v_{0} is 20 \frac {m}{s}

    s_{0} +v_{0}*(0) -\frac{1}{4}(0^2) = 0 gives s_{0} =0

    and

    v_{0} -\frac{1}{2}(0) = 0 gives v_{0} =0

    and solve s(t) to get the distance using v_{0} = 20


    20 -\frac{1}{2}t\Rightarrow t =40

    s(t) = 0 + 20(40) -\frac {1}{4}(40^2) = 400

    thus the answer is 400 metres.

    ps. teaching myself LaTex
    Stopping distance is s(t) - s0 = d = v0 - (a/2)t^2

    v0 is not zero, it is given 20 m/s

    also you need to know v(t) - v0 = - a*t what is v(t)? it is given in the statement of the problem.
    Does this help
    Last edited by votan; October 11th 2013 at 05:36 PM.
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    Re: velocity help

    okay so

    s(t) = s_{0} + v_{0} -\frac{1}{4}(t^2) \Rightarrow s(t) - s_{0} = d = v_{0} - \frac{1}{4}(t^2)

    and v(t) - v_{0} = - a(t) ??
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    Re: velocity help

    Quote Originally Posted by Jonroberts74 View Post
    ps. teaching myself LaTex
    Nice job on the LaTeX.

    Did you copy the problem word for word? Some of this just doesn't make sense.
    Quote Originally Posted by Jonroberts74 View Post
    starting at time t=0, a train is halted by a braking deceleration of \frac{1}{2}\frac{m}{s^2}, with the result that its position at time t is s(t) = s_{0} +v_{0}t -\frac{1}{4}(t^2)

    show that s_{0} is the train's postion when the brakes are applied and v_{0} is its velocity at that time. Find the stopping distances if v_{0} is 20 \frac {m}{s}
    I am very confused about the first two questions. You didn't get them right, at least the way the question is worded. s_0 = 0 can be defined at the starting point, but that was never mentioned, and v_0 had better not be 0 else the train isn't moving at t = 0.

    The brakes are applied at t = 0. I'm not sure what they wanted out of the first question...we need to show that the position of the train at t = 0 ( s_0) is the train's position. So we have
    s(0) = s_0 + v_0(0) - \frac{1}{4}(0)^2 = s_0 So s(0) = s_0 is the position of the train when t = 0.
    (The acceleration is opposite to the speed of the train. Good job in picking up the negative there.)
    And v_0 is equal to v at t = 0. Well, we have
    \frac{ds}{dt} = v(t) = v_0 - \frac{1}{2}t

    at t = 0:
    v(0) = v_0 - \frac{1}{2}(0) = v_0 So v(0) = v_0 is the speed of the train at t = 0.

    Why were these questions even asked?? s(0) = s_0 and v(0) = v_0 are from the definitions of s(0) and v(0). Again we can usually assume s_0 = 0, but this was never done. I just don't get it.

    For the last part we know that v(t) = 0 at the final point, so from the speed equation we have
    0 = 20 - \frac{1}{2}t \implies t = 40, as you said, but:

    s(40) = s_0 + 20(40) - \frac{1}{2}(40)^2

    What is s_0? This was not given in the problem statement. (Typically we set s_0 = 0 by choice of the origin, but this was never mentioned for this problem.)

    Tell your professor for me that you need to set an origin for Physics problems (which is why I'm repeating that.) If you take the train to be at the origin then we have s(0) = s_0 = 0 then the last question can be done.

    -Dan

    Edit: We cross posted. Using s(40) - s_0 = d is a good move. Thanks votan.
    Last edited by topsquark; October 11th 2013 at 06:01 PM.
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    Re: velocity help

    and

    v(t) = 20 - \frac{1}{2}(t), t=0 \Rightarrow v(t) = 20
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    Re: velocity help

    Quote Originally Posted by Jonroberts74 View Post
    and

    v(t) = 20 - \frac{1}{2}(t), t=0 \Rightarrow v(t) = 20
    You had that right in the first post. The train has stopped at some time t. You are trying to find t, so 0 = 20 - (1/2)t, etc.

    -Dan
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    Re: velocity help

    yeah that is word for word the problem.
    yes it is very vague, I had assumed it was referring to s_0 as being the origin.

    thank you!
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    Re: velocity help

    Quote Originally Posted by Jonroberts74 View Post
    yeah that is word for word the problem.
    yes it is very vague, I had assumed it was referring to s_0 as being the origin.

    thank you!
    Perhaps I was a bit manic over that. But I have a real pet peeve when most Calculus teachers try to teach kinematics problems. Their terminology is typically awful, as are their assumptions...they rarely define the origin, much less define positive directions. They simply assume one and never say.

    -Dan
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    Re: velocity help

    Yes the statement of the problem is poor and confusing to most students. I picked up what the question was about from the statement - find the stopping distance. I don't know if this this a textbook question, or the instructor lousy excess of zeal to complicate the life of his students. He could be a lab technician in charge of the course, and that shows how much he has to learn before leading a class on this subject.
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  10. #10
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    Re: velocity help

    Unfortunately, I do not have a professor. I am teaching myself math in the hopes to get into university to study the maths. It can be frustrating but I am persistent. I do see the error in the textbook. I really like differential geometry, I read about it but still a lot of it is beyond my current knowledge base--I will get there.
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