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Find the tangent line that passes through the origin

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data

    This problem is 2.95 of University Physics, 11th edition.

    Catching the Bus: A student is running at her top speed of 5.0 m/s to catch a bus, which is stopped at the bus stop. When the student is still 40.0 m from the bus, it starts to pull away, moving with a constant acceleration of 0.170 m/s2.

    There are subproblems a,b,c,d,e, and f; which I've all figured out except for f)

    f) What is the minimum speed the student must have to just catch up with the bus?

    2. Relevant equations

    Let the subscript b mean bus, and let the subscript p mean student (p for pupil).

    sp(0)=0 m
    sb(0)=40 m

    vp(t)=Unknown

    The acceleration of the bus was given, I used calculus to find the velocity and position.
    sb(t)=0.085t2+40 m
    vb(t)=0.17t m/s
    ab(t)=0.17 m/s2

    3. The attempt at a solution

    I modeled the problem by considering the bus and the student as point particles. The points in time where the student and the bus are at the same place are the intersections of the graphs of the position functions. The position function of the bus is known and given above, but the position function of the student is the integral of the student's velocity, which is constant.

    The position function would be of the form sp(t)=v(t)*t+0

    I managed to figure out that sp(t) must be the equation of the tangent line of sb(t) which passes through the origin.

    How do I find the equation of the tangent line of sb(t)=0.085t2+40 m (And thus, the velocity) that passes through the origin?
     
  2. jcsd
  3. Jun 20, 2012 #2
    Hmm, I didn't have to use calculus to solve this question. Set the displacements of the student and bus equal, then work from there.
     
  4. Jun 20, 2012 #3
    I don't know the displacement without knowing the velocity...
     
  5. Jun 20, 2012 #4
    I mean, take the equations of the displacement of the student and the bus, then set them equal.
     
  6. Jun 20, 2012 #5
    You mean like so?:
    sp(t)=sb(t)

    -> v(t)*t=0.085t2+40,
    -> v(t) =(0.085t2+40)/t


    This tells me what the velocity would be at my intersection point, but I still don't have the time, t.
     
  7. Jun 20, 2012 #6
    Yes, now what kind of equation do you have from there?
     
  8. Jun 20, 2012 #7
    It's a rational equation that gives me the velocity of the student.. If I set that equal to the velocity function of the bus, I can find the time where they intersect!

    0.17t=(0.085t2+40)/t

    t=21.7 s.. Thank you so much!!!
     
  9. Jun 20, 2012 #8
    Ha, I was actually going with a quadratic equation and setting √(v2 - 4ad) = 0, but your method works as well. =)
     
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