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Finding Distance Given Velocity and Time

  1. Aug 31, 2011 #1
    1. The problem statement, all variables and given/known data
    A car accelerates uniformly and reaches a speed of 4.40 m/s in 8.00 seconds. Calculate the distance traveled by the car from a time of 1.40 to 4.70 seconds.


    2. Relevant equations
    x = xo + vo t + ½ a t^2

    (v-vo)/t=a


    3. The attempt at a solution
    (v-vo)/t=a
    (4.40 m/s - 0 m/s)/8.00s=a
    .550 m/s^2=a

    x = xo + vo t + ½ a t^2
    x= 0 + 0(3.30s) + 1/2 (.550 m/s^2) (3.30s)^2
    x= 3.00 meters

    My homework is on LON-CAPA, an online program and when I enter 3.00 meters it states the answer is incorrect. I've been working on this problem for 2 hours and can't come up with any other answer.
     
  2. jcsd
  3. Aug 31, 2011 #2

    Hootenanny

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    Is the velocity really zero at t=1.40 seconds?
     
  4. Aug 31, 2011 #3

    PeterO

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    Looks like you have calculated where the car will be at time 3.30s to me??

    That is not what you wanted - not what you thought you were finding.
     
  5. Aug 31, 2011 #4
    My logic for that is if acceleration is constant, the rate of change of velocity will be constant so you'd cover the same distance from time 0s to 3.30s as you would 1.40s to 4.70s.

    Is at least the acceleration I found correct?
     
  6. Aug 31, 2011 #5

    Hootenanny

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    That would be true if the velocity was constant, which sadly it isn't.
    Yes, your acceleration is indeed correct.

    So, in order to correct your formulae, you need to work out the velocity at t=1.40 seconds.
     
  7. Aug 31, 2011 #6
    I guess I mistakenly thought it was constant because it increases at a constant rate. So in physics, does constant only mean a value stays the same during a period of time?

    I'll rework this and post what I get soon. Thank you so much for the help.
     
  8. Aug 31, 2011 #7
    I think you should think of this as the displacement and not absolute distance.
     
  9. Aug 31, 2011 #8

    Hootenanny

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    In a manner of speaking, yes. In Physics, constant means unchanging. Otherwise, how would we differentiate between something that remained constant, or changed at a constant rate :wink:
    No problem!
     
  10. Aug 31, 2011 #9
    v(1.40s)=.770 m/s

    x = xo + vo t + ½ a t^2
    x= 0 + (.770 m/s) (3.30s) + 1/2 (.55 m/s^2) (3.30s)
    x= 5.54 m
     
  11. Aug 31, 2011 #10

    Hootenanny

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    Looks okay to me :approve:

    (Although there is a small typo in the second line - it should be (3.30s)^2)
     
  12. Aug 31, 2011 #11
    You're right, I slacked off with that for this problem since it all occurs in one direction and went with magnitude alone for position, but that's inconsistent since I used a vector quantity for change in position. I'll keep this in mind for future problems, thank you.
     
  13. Aug 31, 2011 #12

    PeterO

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    Not only in Physics.

    You can sit in a parked car for 10 minutes and your position, velocity and acceleration will all be constant, ie zero.

    You can sit in a car for 10 minutes that is being driven at 20 miles an hour and your speed and acceleration will be constant [one of them will be zero] but your position will certainly change.

    You can sit in a car for 30 seconds that is easing away from traffic lights accelerating at a constant 2 miles per hour, per second.
    All the while your acceleration will be constant, but your speed sure is changing - you will be doing 60 mph at the end of the 30 seconds!
    And your position will be changing as well.

    EDIT: I used miles per hour since the speedometer in a car is not calibrated to metres per second. Change miles to kilometres if you are from a metric country.
     
    Last edited: Aug 31, 2011
  14. Aug 31, 2011 #13
    Whoops, nice catch. I see where I went wrong in my logic now. I made a graph of velocity over time and the area under the line from 0s to 3.3s is much smaller than the area under the line from 1.40s to 4.70s. The only way to make what I originally did work is if, like you said, velocity was constant, making both these areas under the line equivalent.

    Thanks again for the help, everyone on this forum is always so kind and I really appreciate people like you who use their free time to help those struggling with this subject.
     
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