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Help with an acceleration and distance problem

  1. Sep 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I asked a previous question that someone answered. They said to use two equations which i did, but it still didn't work. The problem is: a rocket sled goes 632 mil/hr and slows down to a stop in 1.40 seconds. find the negative acceleration and find the distance traveled during this negative acceleration. The equations suggested were: v = u + at for acceleration and v^2= u^2 + 2aS. I know the S means distance, and u stands for initial velocity. i converted the mil/hr --> met/sec


    2. Relevant equations
    i attempted to use the equations given: v = u + at and v^2= u^2 + 2aS to find the acceleration and distance and i also attempted to use the distance formula x = x initial + v initial X t + 1/2 at^2 to try to find the distance, which didnt work.
    My book answers were -202 met/sec^2 acceleration and 119 met distance. i understand why acceleration is negative but don't know what i am doing wrong as far as solving.


    3. The attempt at a solution
    632 mil/hr --> 926.9 met/sec
    632 mil/hr X 5280 ft/1 mil X 1 hr/3600 sec = 926.9 met/sec

    using v = u + at = 926.9 met/sec + a1.40 sec. = -926.9met/sec / 1.40 =
    -662.0 met/sec^2 acceleration??! is not even close to -202 met/sec.^2

    using v^2 = u^2 + 2aS = (926.9 met/sec)^2 + 2(-662.0met/^2)S =
    -859143.6 = 1324S= -859143.6/-1324 = 648.9 meters?? no where close to 119 meters.

    using x = x initial + v initial X t + 1/2 at^2 = 0m + 926.9 met/sec. X 1.40 sec. + 1/2a(1.40)^2 = 1,297.6 + 1/2a1.96= -1,297.6 = 0.98 = -1,324 met/sec^2
    A different answer from -662 met/sec^2 but still wrong.

    using a = change in veloc. divided by change in time = 0 met/sec - 926.9 met/sec. divided by 1.40 sec. = -662.0 met/sec^2. same answer from using the v = u +at equation, and still wrong. there is something i'm not doing right, or something i'm missing, or both. i am lost as to what to do now. please help me understand step by step what i should be doing to solve these problems correctly.
     
  2. jcsd
  3. Sep 17, 2008 #2

    danago

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    Gold Member

    Is your conversion to m/s correct? I dont ever use the imperial measurement system so i dont really know the conversion factors, but according to an online converter i just used, you havent converter correctly. Try that again.

    EDIT: I just realised, you converted to ft/sec not m/s. Convert the miles to METERS not FEET.
     
  4. Sep 17, 2008 #3
    :0 Oh!!! how stupid of me!!! i didnt notice what i did. i'll try and see if it works out correctly
     
  5. Sep 17, 2008 #4
    ok, i converted it correctly:
    632 mil/hr X 1609.34 met/mil X 1 hr/3600 sec = 282.52 met/sec
    then i used v = u + at
    v = u + at = 282.52 + a1.40sec^2 = -282.52 / 1.40 = -202 met/sec^2 acceleration which is correct. but when i tried using v^2 = u^2 + 2aS it didn't work:
    v^2 = u^2 +2aS = 282.52^2 + 2(-202)S = 79817 + 2(-202)S = -79817 = -404S=
    -79817/-404 = 197 meters. this isn't right. the correct answer should be 119 meters.
     
  6. Sep 17, 2008 #5
    even using the distance formula: x = xinitial + vinitial t + 1/2 at^2 yields 197.5 meters which isnt the correct answer for distance. arrrrghhh!!! so i have the correct acceleration now but not the correct distance. hmmm.
     
  7. Sep 17, 2008 #6

    LowlyPion

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    Homework Helper

    You can use the V2 = 2*a*x
    x = (282.52)2/2*202 = 197.5

    Are you sure you have the right initial conditions?

    119 implies an acceleration in 1.4 sec of 121.42 and an original velocity of 170 m/s
     
  8. Sep 18, 2008 #7
    i know, but 119 meters is the answer my book gave me for the total distance. there can be typos but they aren't very often. i'll keep trying to calculate. :rolleyes:
     
  9. Sep 18, 2008 #8

    LowlyPion

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    Homework Helper

    Unless there are some other conditions not mentioned by the problem statement, trust the math.

    Good luck.
     
  10. Sep 18, 2008 #9
    My mistake!! My answer is 198 meters! i looked at the answer in the book again and it's the same. i dont know why i kept thinking it was 119 meters! Math doesn't lie! i did do it correctly :)
     
  11. Sep 18, 2008 #10
    Solved :)
     
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