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Relative motion on a bike

  1. Sep 21, 2015 #1
    1. The problem statement, all variables and given/known data

    You are travelling on a bike, moving at a constant speed of 20,0 km/h. You just finished eating an apple, and you wish to pop the core into a trash bin located by a busstop.The bin is 3,00 m from the edge of the road (where you are biking). You throw it at a point 4,00 m before you pass the trash bin. The apple core behaves as a projectile under the force of gravity g = 9,80 m/s^2, and you can ignore friction and wind. The height of the (top of the) bin is the same as the height of your hand as you throw, 1,00
    m above the ground.

    upload_2015-9-21_12-13-17.png

    a) You throw the apple upwards at an angle of = 20 degrees with horizontal. With what initial velocity vector relative to the Earth should you throw the apple to hit the bin?

    b) With what initial velocity vector relative to you and the bike should you throw the apple to hit the bin? What speed does that correspond to?

    c) As it happens, a little old lady is standing exactly halfway between you and the bin. She is 160 cm tall. Do you hit her? Remember to draw a sketch of the situation.

    2. Relevant equations
    For (a) I used the range equation, R=(v^2 * sin(thetha))/g

    3. The attempt at a solution
    For (a) I used the range equation R=(v^2 * sin(thetha))/g and got 8,73 m/s.

    In (b) I struggle with getting started. If the bike relative to the earth is the velocity vector in y-direction (20 km/h), and the the apple relative to earth is the second vector. Then the velocity vector for me relative to the bike is the opposite vector from the angle? I am really stuck on this one, hope someone can help me here:)
     
  2. jcsd
  3. Sep 21, 2015 #2

    RUber

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    Could you solve the problem if the bike was not moving? If you can, just subtract the velocity of the bike from your stationary solution.
     
  4. Sep 21, 2015 #3

    RUber

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    I concur that the total velocity in the direction of travel is what you found for part (a), |V|=8.73m/sec.
    What portion of that is in the +z direction? What portion is in the xy plane? Of the portion in the xy plane, what are the x and y components?
    Subtract off the velocity of the bike from the y component and you should have a solution.

    The question about hitting the lady can be reduced to the 1-dimensional problem based on the initial upward velocity and time to the middle of the path.
     
  5. Sep 22, 2015 #4
    how did you get 8,73 m/s on a)?
     
  6. Sep 22, 2015 #5
    Have you managed to do b or c?
     
  7. Sep 22, 2015 #6

    RUber

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    I agree that the range equation as listed :
    Does not lend itself right away to finding 8.73 m/sec.

    However, the range equation can be derived by using two pieces of information:
    1) The total change in z is zero. ##R_z = vt_f\sin\theta - \frac{g}{2}t_f^2 = 0##.
    Solving for t, you get the trivial solution of t=0, or ##t_f = \frac{2v}{g}\sin \theta##
    2) The total distance in the x-y plane ## 5m = R_{xy}= v t_f \cos\theta ##. Replacing t_f with what you found above gives:
    ## 5m =v\frac{2v}{g}\sin \theta \cos\theta =\frac{2v^2}{g}\sin \theta \cos\theta ##
    Rearranging to solve for v, you get:
    ## v = \sqrt{ \frac{5g}{2 \sin \theta \cos \theta } } ##
    Which gives v = 8.731m/sec.
     
  8. Sep 22, 2015 #7
    I got that v = sqrt((g*5)/(2(tan(theta)*cos^2(theta)))
    which also gives 8.73m/s, do you think this is the correct answear then?
     
  9. Sep 22, 2015 #8

    RUber

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    tan x cos^2 = sin x cos.
     
  10. Sep 22, 2015 #9
    May in b it is just that simple as to covert 20km/h to m/s, then subtract the result from the answer in a?
     
  11. Sep 23, 2015 #10
    So you think if the velocity should add up to be 8,73m/s as if the bike was stationary relative to the earth, the velocity relative to you and the bike should be 8,73=(20km/h)/3.6 * v(youandbike), and solve for v(youandbike) = 3.17m/s ?
     
  12. Sep 23, 2015 #11
    In a), dont you need an angle in your answer because they are asking for a vector? know its relative to the earth, therefore i am not exactly sure if you need an angle in your answer
     
  13. Sep 23, 2015 #12

    RUber

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    I assume that the problem is asking for the vector in component form.
    ## \vec V = V_x \hat x + V_y \hat y+V_z \hat z##.
    The velocity of the bike is pointed in my +y direction, so in vector form,
    ## \vec V_{bike} = 0 \hat x + V_{bike} \hat y+0 \hat z##, where
    ##V_{bike}= | \vec V_{bike} |##
    For part b, the vector describing how you throw the apple is exactly ##\vec V - \vec V_{bike}##
    But the magnitude of that vector is not ##|\vec V| - |\vec V_{bike}|##, since you can only directly subtract magnitudes if the vectors are pointing in the same direction.
     
  14. Sep 23, 2015 #13
    Is this correct method on b?
    From bicycle:
    Vxi=5.56 m/s

    From throw
    Vxi=Vi*cos(36.7)
    Vyi=Vi*sin(20)
    Vzi=Vi*cos(53.3)

    Then i find the velocity in yz-direction using Vyi=1/2*g*t, where t=5/(Vzi*cos(53.3)), and velocity in xy-direction using Vyi=1/2*g*t, where t=5/(Vxi*sin(36.7)). Add the Vyi-velocities and subracts the velocity from bicycle in Vxi-direction.
    And then i take sqrt(Vxi2+Vyi2+Vzi2). That makes me end up with 9.77 m/s.
     
  15. Sep 23, 2015 #14

    RUber

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    Nope.
    You need to keep a cos(20) term in your Vx and Vz numbers. Otherwise your sum of squared components will not add up to Vi^2.
     
  16. Sep 23, 2015 #15
    How did you get the angles 53.5 and 36.7?
     
  17. Sep 23, 2015 #16
    36.7 is the angle between where you start throwing and the red arrow. 53.3 is 90-36.7. But should probably not use them.
     
  18. Sep 23, 2015 #17

    RUber

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    The angles are fine. Notice that you can keep a more precise answer by saying that the right/left component (x in my diagram) is 3/5 the total 2d velocity, and the front/back component (y in my diagram) is 4/5 the total 2d velocity.
    Otherwise, cos(53.3) = .5976, cos(36.7) = .8018. These are probably close enough to not affect your rounding...but why go through the effort of finding the angles when the side lengths are given to you and the hypotenuse is simple enough to figure out?
     
  19. Sep 23, 2015 #18
    Vxi=Vi*cos(36.7)*cos(20)=6.563 m/s
    Vyi=Vi*sin(20) = 2.986 m/s
    Vzi=Vi*cos(53.3)*cos(20)= 4.922 m/s

    Vxrel=Vxi - Vbike= 6.563 - 50/9 = 1.007 m/s

    Virel = Vxrel / cos(36.7)*cos(20)= 1.007/ cos(36.7)*cos(20)= 1.18 m/s

    is 1.18 m/s the initial velocity vector relative to you and the bike should you throw the apple to hit the bin?
     
  20. Sep 23, 2015 #19

    RUber

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    1.18 m/sec is a speed. And no.
    I agree with what you did up to ...
    there.
    But remember, your throw is responsible for all of the other components.
     
  21. Sep 23, 2015 #20
    Looks good. For Virel, check your answer by first calculating time of flight, t, and see if 1.18m/s "gets you there."
     
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