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Homework Help: How do I use physics equations for this?

  1. Feb 2, 2012 #1
    1. The problem statement, all variables and given/known data

    a vehicle starts from rest and accelerates at a reate of 2.0m/s2 in a straight line til it reaches a speed of 20m/s. then it slows to constant 1.0m/s2 til it stops.
    how much time elapses from start to stop?
    how far does the vehicle go?


    2. Relevant equations
    I have:
    a1 = 2.0m/s2
    a2 = 1.0m/s2
    v1 = 2.0m/2
    v2 = 0.0m/2


    3. The attempt at a solution

    what I did was this:
    t1=0s then 2m/s2 * 0s= 0
    t1=1s then 2m/s2 * 1s = 2m/s
    t1=10s then 2m/s2 * 10s = 20m/s

    that gives me 10 seconds later the vehicle gets to a = 20m/s

    so now the same thing for slowing down setting my 20m/s to t=0 and working from there
    t2=0 then 1m/s2 * 0s = 0
    t2=1 then 1m/s2 * 1s = 1m/s
    t2=20 then 1m/s2 * 20s = 20m/s

    now I have:
    total time: t2 + t1 = 20s + 10s = 30s

    this is the only equation given that I know how to apply:
    total distance: based on a = d/t set to solve for d so d = at
    total distance: d1 = 2.0m/s * 10s = 20m
    d2 = 1.0m/s * 20s = 20m
    20m + 20m = 40m

    total time travelled is 30s
    total distance travelled is 40m

    the problem is I did not use any of the basic equations I was given in my book for this. I am at a loss as to how to match my equations to my work...... and seeing as this is the first week of the semester I have a feeling this is going to come back to haunt me drastically in about 4 weeks.

    thanks for any advice offered.
     
  2. jcsd
  3. Feb 2, 2012 #2

    tiny-tim

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    hi netrunnr! :smile:
    yes, that's correct, but using v = vo + at would have been a lot simpler and quicker …

    that is what physics equations are for!
    no, a = d/t is wrong :redface:

    look up your constant acceleration equations, and choose one that seems appropriate to find the distance(s) …

    show us what you get :smile:
     
  4. Feb 2, 2012 #3
     
  5. Feb 2, 2012 #4

    tiny-tim

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    ok so far :smile:
    i don't understand any of this :redface:

    is r meant to be speed? (if so, please use v, like everyone else :wink:)

    you've written r = d/t (instead of a), but you've used the acceleration anyway
     
  6. Feb 2, 2012 #5
    Hmmm I seem to have my vocabulary messed up....
    Would it be correct to say
    a = distance/time?
     
    Last edited: Feb 2, 2012
  7. Feb 3, 2012 #6

    tiny-tim

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    (just got up :zzz: …)
    a is for acceleration
     
  8. Feb 3, 2012 #7
    So what exactly is the rate that distance divided by time gives? Is it not acceleration?
     
  9. Feb 3, 2012 #8

    tiny-tim

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    speed!! :smile:

    that's why speed is measured in m/s ! :rolleyes:

    (and of course, only if the acceleration is zero)

    get some sleep! :zzz:​
     
  10. Feb 3, 2012 #9
    omg a *duh* moment indeed.
    lesson here : do not overthink a solution :biggrin:
     
  11. Feb 9, 2012 #10
    just learned something very important today regarding this - and it was the source of my confusion here:
    Speed is the rate of change of distance with time.
    Velocity is the rate of change of displacement with time.

    I thought speed was acceleration.... so on that note:

    my problem is a total time of 30s
    and the distance traveled is:
    x = xo + vot = 1/2at2
    so part one is x1=0 + 0 + 1/2*2*102 = 100m
    then part 2 is xtotal=100 + 20m/s+1/2*(-1)*202 = 300m

    so the total distance traveled is 300m! not 30.

    is this right?
     
  12. Feb 10, 2012 #11

    tiny-tim

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    hi netrunnr! :smile:
    ah, that explains a lot!

    i was getting worried! :biggrin:
    yes, that's fine! :smile:

    (warning: in an exam, don't do two lines in one, as in your last line …

    there's nothing wrong with it, but there's so much chance of making a silly mistake that it's not worth it :wink:)

    btw, you can also get there without using t, by using v2 = u2 + 2as (twice) …

    try it! :smile:
     
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