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Confused with simple equation

  1. Jul 1, 2008 #1
    1. Can someone please help with this equation



    2.An accelerating lab cart passes through two photo gate timers 3.0 m apart in 4.2 s. The velocity of the cart at the second timer is 1.2m. What is the cat's velocity at the first gate?



    3. So I am so confused as to how to find the initial velocity. I used d=(Vi + Vf/2) *t which is 3.0= (Vi + 1.2/2) *4.2. The solution Vi=0.23m/s. I have no clue if I am way off. Can someone please tell me if I am on the right track. Thank you!

    The last time I took physics was 8 years ago and now I'm trying to upgrade. Wish I spent more time on homework, less on hair!
     
  2. jcsd
  3. Jul 1, 2008 #2

    malawi_glenn

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    "The velocity of the cart at the second timer is 1.2m"

    you mean 1.2m/s here

    and isn't this the formula which you should use?

    d=(Vi + Vf) *t /2 ?
     
  4. Jul 1, 2008 #3
    So the equation is wrong? That's silly, I took it off of the website I was provided with through school. Go figure. So the answer is initial velocity is -0.843m/s. Is this right?
     
  5. Jul 1, 2008 #4
    Thanks for your help malawi_glenn, I feel like the silly girl sitting in the room full of math geniuses all over again.
     
  6. Jul 1, 2008 #5

    malawi_glenn

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    You think it is plausible that the initial velocity has opposite sign?

    And I made that question so that you think if the equation you used is correct, why should the first equation you used be correct?

    What does (Vi + Vf)/2 represent? It represent the mean velocity over that perticular distance and time.

    I would really love to see how you got V_i = -0.843m/s

    Do you know how to use a calculator? It does not work the same way as we calculate things on the paper, just so you are aware of that.
     
  7. Jul 1, 2008 #6
    My, what a humbling comment. Right then, I'll go back to the equation I had originally used and I'll try to work it out.

    3.0m=v_i + 1.2m/s/2 * 4.2s

    3.0m/ 4.2s= 0.714

    0.714-1.2=v_i/2

    -0.486 x 2 = v_i

    v_i = 0.97m/s

    Should I just give up now?
     
  8. Jul 1, 2008 #7

    alphysicist

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    Hi Kristin,

    As malawi_glenn pointed out, your equation is wrong. The correct equation is:

    [tex]
    d = (v_i + v_f)\frac{t}{2}
    [/tex]

    where d is the displacement. This does not give an answer of -.843 m/s for [itex]v_i[/itex]; I get the correct answer stated in your original post. Try it again, and if you still get the same wrong answer, please post the steps you did in your calculation so we can see what you did.
     
  9. Jul 1, 2008 #8
    Thank you
    I tried the equation above and got an answer of Vi=0.23m/s
    I am doing this course via distant education, so there is not much support. The print out they supplied did not even include the equation I needed to figure out this problem so I was searching online.
    I really appreciate your help.
     
  10. Jul 1, 2008 #9

    alphysicist

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    Many courses seem to focus on the following three kinematic equations:

    [tex]
    \begin{align}
    v&=v_o + a t\nonumber\\
    d &= v_o t + \frac{1}{2} a t^2\nonumber\\
    v^2 &= v_o^2 + 2 a d \nonumber
    \end{align}
    [/tex]

    with d being the displacement. If that's what you are given, then you can solve this problem by solving the first equation for [itex]a[/itex], and then plugging it into the second equation. That will give you the equation you used above.
     
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