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Uniform acceleration vector problem

  1. Nov 9, 2005 #1


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    i just need some1 to tell me if i did something wrong and if i did a hint in the right direction would be great.

    an object accelerates uniformly at 8 m/s^2, 130 degrees for 5 seconds until it reaches a velocity of 20 m/s, 143 degrees.
    Calculate: a) the intial velocity b) the displacement at the end of the 5 seconds

    a) i decided i was gonna use [tex]a=\frac{v_2-v_1}{\Delta t}[/tex]
    so i had to turn the vectors to rectangular form, so i worked out their components using trig:

    8, 130 degrees = [-5.14, 6.13]
    20, 143 degrees = [-16, 12]

    so i filled in wat i know:

    [tex][-5.14, 6.13] =\frac{[-16, 12] - [x,y]}{5}[/tex]

    [tex]5*[-5.14, 6.13]=[-16, 12] - [x,y][/tex]

    [tex][-25.7, 30.7]-[-16,12]=-[x,y][/tex]

    [tex]\frac{[-9.7, 18.7]}{-1}=[x,y][/tex]

    [tex][9.7, -18.7]=[x,y][/tex]

    so know i got the components of the intial velocity, to put them in polar form:

    [tex]v_1=\sqrt{443.78} \sim 21.1 m/s[/tex]

    using tan inverse i get 27.4 degrees for the angle and because x is negative and y is positive you get 21.1 m/s , West 27.4 degrees North
    that wat i get for "a"

    b) i chose to use [tex]\Delta d= v_2 \Delta t - \frac{1}{2}a \Delta t^2[/tex]

    [tex]\Delta d= [-16, 12]*5- \frac{1}{2} [-5.14, 6.13]*25[/tex]

    [tex]\Delta d= [-80,60]-[-64.25, 76.63][/tex]

    [tex]\Deta d=[-15.75, -16.63][/tex]

    Delta d= ~ 23 m W 46.5 degrees S

    howd i do? im not so confident
  2. jcsd
  3. Nov 9, 2005 #2
    I haven't got a calculator with me, but assuming you typed it in right, I agree with you answer to (a). However, your equation for displacement has to use initial velocity (whereas you used final velocity), you can prove this with a velocity-time graph. If you want the graph tell me and I'll add it on in the morning (UK time).

    d = V(initial) x T + 0.5 x A x T^2 (sorry about the crude formula)

    I think you should be more confident, your thinking is spot on (especially how you approach with questions, just need to learn the formulas) ;-)

  4. Nov 9, 2005 #3


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    my apologies about the thread being posted three times, i dont know why but everytime i edited the thread it just posted a new 1, sorry.

    thanks for the reply BerryBoy, ill apply that formula, dont worry bout the graph, thanks for the offer though :)
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