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Kinematics in 2D

  1. Oct 3, 2003 #1
    ok so here are some problems that I need help with.. I will explain how I did them, and if anyone can tell me why I am getting them wrong and what I should do to fix it would be great.. Thanks...
    1.)A wild horse starts from rest and runs in a straight line 29 degrees north of west. After 34s of running in this direction, the horse has a speecd of 17m=m/s..
    a.) north and east are positive lines, fine the component of the horses acceleration pointing along the north south line.

    Here is what I did.. I found the average acceleration to be .5m/s^2.. then using the value .5 and the time 34s.. I plugged it into the equation Vxf=Vxo+AxTf.. so it would be Vxf=0+.5(34) which is 17..
    b) find the component of the horses acceleration that points along the east west line..
    I used the same equation but For Vo I used 17.. and used the rest of the numbers at before.. and I got..34.

    Q2) When chasing a hare along flat ground,a grayhound leaps into the air at a speed of 12.3m/s at an angle 27degrees above the horizontal.
    A) what is the range of his leap?
    To do this I found, T setting 5.6(found from 12.3(sin27)) added to 1/2(-9.80)T equal to zero. The time comes out to 1.1s, then I used this multipiled by 12.3cos27.. to find the range... I just dont know if I did this right...

    Q3) Car drives straight off a cliff 55m High, the impact of the car is 140m from the base of the cliff... how fast was the car traveling when it went over the cliff..
    To do this one,I found the time by Using the Equation T=the sq.root of 2H/g.. so The sq.root of 2(55)/9.80.. to equal 3.4. then using the time, I used the equation Vyf=-g(T) so -9.80(3.4) which equals -33.3


  2. jcsd
  3. Oct 3, 2003 #2


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    In problem 1 you are asked to find the component of acceleration.
    The formula you are using "Vxf=Vxo+AxTf" is for speed.
    Since you know that the average acceleration is 0.5 m/s2,
    at 29 degrees north of west, the north component is -0.5 cos(29) and the west component is -0.5 sin(29).

    The y component of position is 5.6T- (1/2)(9.8)T2= 0 which has two solutions: T= 0 (of course) and T= (2*5.6)/9.8= 1.1 seconds. The x coordinate is (12.3 cos(27)T= 12.5 meters. You copied the formula wrong here ("T" instead of "T2) but that looks like what you are doing.

    Assuming the car drove horizontally off the cliff at speed v0, it's horizontal position will vo T and its vertical position will be
    55- (1/2)(9.8)T2. It hits when 55- (1/2)(9.8)T2= 0 or T= [sqrt](2*55/9.8), atain what you have.
    However, "Vyf= -g(T)" appears to be the vertical speed of the car as it hit the base of the cliff. You were asked for the speed of the car "when it went over the cliff"- its initial speed.
    Since you now know T= 3.4 seconds and vo T= 140, you can solve for
    v0= 140/T= 140/3.4= 41.4 m/s.
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