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SIMPLE QUESTION, NEED HELP!

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  1. Aug 13, 2015 #1
    • Warning: Thread titles should be descriptive of the thread content, not a plea for help.
    1. The problem statement, all variables and given/known data
    An airplane accelerates down a runway at 3.20 m/s2 for 32.8 s until is finally lifts off the ground. Determine the distance traveled before takeoff.

    2. Relevant equations


    3. The attempt at a solution

    Okay, so i get how to write it, and I understand what the majority of the equation means. But when I do the actual math, the answer is totally effed up. Haha. How am I calculating wrong? Could someone please break down the calculations part for me so I can better understand this kind of question?

    d = vi*t + 0.5*a*t2

    d = (0 m/s)*(32.8 s)+ 0.5*(3.20 m/s2)*(32.8 s)2


    The answer in the answer key says it is:

    1720 M

    ..that is not what I got.
     
  2. jcsd
  3. Aug 13, 2015 #2

    e.bar.goum

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    What did you get?

    Are you aware that "2" at the end of your equation should be squared?

    that is,

    ##d = v_i t + 0.5 a t^2##

    I get the same answer as the key using that equation.
     
  4. Aug 13, 2015 #3

    billy_joule

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  5. Aug 13, 2015 #4

    Ahh yes. I am aware is should be squared. I'm completely new to physics and am struggling a bit. Is the first part of the equation: d = (0 m/s)*(32.8 s)
    not equal to zero since you are multiplying 32.8 by zero? From there i'm just..sort of lost. I apologize if this seem ignorant and basic.
     
  6. Aug 13, 2015 #5
    Haha my ignorance is killing me. I guess i'm just not sure how to go about the math. I'm assuming the first part d = (0 m/s)*(32.8 s) is equal to zero since you are multiplying it...and then what?
     
  7. Aug 13, 2015 #6

    e.bar.goum

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    Yep! That part is absolutely equal to zero. What about 0.5*(3.20 m/s2)*(32.8 s)^2 ?
     
  8. Aug 13, 2015 #7
    So then I assume I take half of 3.20 and multiply it by 32.8 squared?
     
  9. Aug 13, 2015 #8

    e.bar.goum

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    Sure. What do you get?
     
  10. Aug 13, 2015 #9
    By George, I think I've got it!

    I'm not even sure what I was doing wrong now before! Hahah. 1721.344

    Ahh thank you for humoring me on this...I think perhaps I wasn't even squaring at the end. Couldn't even tell you now. :)
     
  11. Aug 13, 2015 #10
    Since there is no initial velocity, ## x = \frac{at^2}{2} ##, you know what the acceleration is, the time is also given and ##\frac{1}{2} ## is ##\frac{1}{2}##, work it out !!
     
  12. Aug 13, 2015 #11

    e.bar.goum

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    Great! And no worries, we've all been there before!
     
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