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Newtons laws

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A person is riding a bicycle at a constant speed v along a straight road and then gradually applies the brakes until the bicycle comes to a stop. Assume that the magnitude of the braking force increases liearly in time acordinto F = bt where b is a constant. The rider & bicycle have a total mass m

    a)what is the magnitude and directio nof teh acceleration of the rider and bicycle

    b) is the acceleration constant or not? Explain

    c) How long does it take to come to a stop

    d) how far does the bike travel during the time the brakes are applied

    2. Relevant equations



    3. The attempt at a solution

    a) i need to find the acceleration of the bike

    v = v0 + at, v = 0

    a = -vo / t

    [tex]\sum[/tex]Fx = ma = -vo/t - bt

    a = (-vo/t - bt)/m

    b) not possitive but i do think it is constant because the brakes are being applied @ a constant, so it is constantly decreasing. does that make sence?

    c) assuming my answer is correct to part a, i would just solve for t, but im not too comfortable with my answer to a so i havent solved it yet

    d) same as c, but i would solve for x
     
  2. jcsd
  3. Oct 5, 2009 #2

    kuruman

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    Part (a) is incorrect. You are given a force F = bt. How is that force related to acceleration?
     
  4. Oct 5, 2009 #3
    Fx = ma = bt

    so a = bt/m

    how do i allow for the initial velocity of the bicle, do i need to use a kinematic equation
     
  5. Oct 5, 2009 #4

    kuruman

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    The kinematic equations are valid only if the acceleration is constant. Here, the acceleration depends on time, therefore you cannot use them. You have enough information to answer parts (a) and (b). For part (c), you need to integrate the acceleration to find the velocity v(t) as a function of time. Once you have that, you can do part (d) by integrating the velocity to find the position x(t) as a function of time.
     
  6. Oct 5, 2009 #5
    a)what is the magnitude and direction of the acerlation ?

    Fx = ma = -bt mag of accel = -bt/m (
    should it be possitive, is the mag the absolute value)

    So the direction is negative

    b)is the accerleration constant?

    No because it is a functio of time

    c)how long does it take to stop

    v = [tex]\int[/tex]a dt = [tex]\int[/tex] (-bt)/m dt

    v = -(bt2)/2m

    t = [tex]\sqrt{2mv/-b}[/tex]

    d)
     
  7. Oct 5, 2009 #6
    d) x = [tex]\int[/tex]vdt = [tex]\int[/tex]-bt2/2m dt

    x = -bt3/6m this is not correct
     
  8. Oct 5, 2009 #7

    kuruman

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    I would say opposite to the velocity instead of negative.
    Good.
    You forgot the integration constant. You want the velocity to be v0 at t = 0, not zero as your expression suggests. Fix that first, then find the time it takes to stop.
     
  9. Oct 5, 2009 #8
    [tex]\int[/tex]v = -bt2/2m + c where c = vo

    t = [tex]\sqrt{2mv_o/b}[/tex]

    part d,

    x = [tex]\int[/tex]-bt2/2m + vodt = -bt3/6m + vot + c

    the answer is x = 2/3vo(2mvo/b).5
     
  10. Oct 5, 2009 #9

    kuruman

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    Looks good.
     
  11. Oct 5, 2009 #10
    i didnt get the same answer as the book (x = 2/3v(2mv/b).5), where did i go wrong
     
  12. Oct 5, 2009 #11

    kuruman

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    I can't tell you where you went wrong unless I see what your answer is and what you did to get it.
     
  13. Oct 5, 2009 #12
    v = -bt2/2m + vo

    so i integrated v to get the position

    x = [tex]\int[/tex] -bt2/2m + vo = -bt3/6m + vot + c


    but the answer is x = 2/3vo(2mvo/b).5
     
  14. Oct 5, 2009 #13
    i attempted to pug in t, but that didnt work out either
     
  15. Oct 5, 2009 #14

    kuruman

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    Call the second integration constant c = 0. You assume that when he starts braking he is at the origin. What you have above is the position at any time t. You want the position when he stops. You know how long it take him to stop, so...
     
  16. Oct 5, 2009 #15
    x = -bt3/6m + vot where t = (2mvo/b).5

    x = -b(2mvo/b)3/2/6m + vo(2mvo/b).5

    im stuck
     
  17. Oct 5, 2009 #16

    kuruman

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    You need to remove the parentheses by raising what's inside to the appropriate powers so you can simplify. For example,

    [tex]b\large(\frac{2mv_{0}}{b} \large)^{3/2}=b\frac{2^{3/2}m^{3/2}v_{0}^{3/2}}{b^{3/2}}[/tex]

    Then the b in the numerator and the b3/2 in the denominator combine to give you a b1/2 in the denominator.
     
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