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Equations for Accelerated Motion (Two problems)

  1. Oct 21, 2013 #1
    1. The problem statement, all variables and given/known data
    A sailboat in a race is moving at 2.5 m/s when it crosses the starting line. It begins to accelerate at 1.1 m/s^2. How long does is take to travel 80m past the starting line?

    2. Relevant equations
    vavg=(vfinal+vinitial)/2
    t=x/vavg
    v^2=vinital+2aΔx

    3. The attempt at a solution
    First I listed all the data I know, stated directly or indirectly in the passage:
    a=1.1 m/s
    x=80m
    vinital=2.5 m/s

    Then I used this formula to find the final velocity:
    v^2=vinital+2aΔx
    v^2=2.5^2 m/s(1.1 m/s^2)*80 m
    v^2=2.5^2 m/s (2.2 m/s^2)*80 m
    v^2=182.25 m^2/s^2
    v=13.5 m/s

    Next I found the average velocity:
    vavg=(vinital+vfinal)/2
    vavg=(2.5+13.5)/2
    vavg=8

    Finally I found the time:

    t=x/vavg
    t= 80m/8 m/s
    t=8s

    Could someone please confirm(or deny) that the time is 8s? I'm somewhat new to the physics world and like to use every resource possible to keep up my GPA. Thank you in advance!

    1. The problem statement, all variables and given/known data
    Many modern cars are designed so hat the front and the back of the car will collapse before the passenger compartment collapses. This allows the front and back of the car to absorb the impact from a collision while making injury to the passenger less likely or less severe. Supposse a car is traveling at 40 mi/h when the driver runs off the roaf and strikes a large tree. The front of the car collapses and the driver comes to a stop ovet a distance of 0.75 m. Find the acceleration of the driver in m/s^2. Also, convert your answer to g's, where 1g=9.8m/s^2


    2. Relevant equations
    v^2+2aΔx

    3. The attempt at a solution

    I started by listing known information:
    vinital=40 mi/h
    vfinal=0
    x=0.75 m
    1g=9.8m/s^2

    Then I plugged the information into my equation:
    v^2=v0^2+2aΔx
    0=1600+2a(0.75)
    0=400=2a(0.74)
    0=400+2a(0.75)
    0+533.3{repating 3}(0.75)
    a=533.3m/s^2

    Then, converting to gs:

    a=533.3m/s^2/9.8m/s^2
    a=54.42g

    Again, would someone please confirm or deny this? :smile:

    For whatever reason I am not so sure of this answer.
     
  2. jcsd
  3. Oct 21, 2013 #2

    haruspex

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    You mean vfinal = vinitial + aΔt
    There are five standard variables in five standard constant acceleration formulae. Each formula relates four of the five. It's worth memorising them. When you need to use one, check which four of the variables your are interested in and select the corresponding equation. In this case you have s, vi and a, and you want to find t. The equation for that is s = vit + at2/2.
     
  4. Oct 21, 2013 #3
    Thank you! I haven't gotten more than a few chapters into physics yet, so I stuck to the formula that my teacher gave. I see how the first formula you gave equals mine. I'm not sure why my teacher gave me the one he did, it seems easier to use yours. Thanks for the tip! Did I give the incorrect answer?
     
  5. Oct 21, 2013 #4

    haruspex

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    I just realised the formula you meant was vf2 = vi2+2as ( s being the same as your Δx). You had left out the square on the right. Sorry for the confusion. So your method was ok, but not the quickest.
    Almost, but you made a mistake at the final step. Don't kick yourself too hard.
     
  6. Oct 21, 2013 #5
    Thanks! Confusion is fine. Confusion seems to be the first thing that leads to great science. :-)


    Is the answer closer to 40.82? I must have divided instead of muliplied.

    What about the first problem? Is that one correct?

    Thanks again!
     
  7. Oct 21, 2013 #6
    I really hate how I can't edit the title. The missing "d" is driving me insane.
     
  8. Oct 21, 2013 #7

    haruspex

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    ah - more confusion. All my replies were in response to your first question. Your answer is not quite right. You went wrong here:
    t= 80m/8 m/s
    t=8s​
    It is better not to post two questions in one thread, for various reasons, unless one is a continuation of the other. I'll take a look at the second question when I can.
     
  9. Oct 21, 2013 #8
    Oh! That's embarrasing. xD Basic arithmetic. 10s. Thanks! My teacher would have had a laugh at that one.

    That's a good idea. I didn't want to be annoying by posting twenty threads in a row, but I supposse that's better than being confusing.
     
  10. Oct 21, 2013 #9

    gneill

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    Ask and ye shall receive. Fixed :smile:
     
  11. Oct 21, 2013 #10
    Haha, thanks, gneill!
     
  12. Oct 21, 2013 #11

    haruspex

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    Is the speed really given in mph? If so you'll need to convert first.
    Did you mean =0+1600+2a(0.75)?
    How did the 1600 become 400? I assume the .74 is a typo.
     
  13. Oct 21, 2013 #12
    Yes, the .74 is a typo. Thanks for pointing that out!!!! I accidentally divided by two, twice. It should be 0=800+a(0.75)

    Multiplying by 75: a=600
    After dividing to convert to gs: a≈61.22

    Is that better?
     
  14. Oct 22, 2013 #13

    haruspex

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    Think that step through again.
     
  15. Oct 22, 2013 #14
    Thank you!

    1. I forgot to do a unit conversion to convert mi/h to m/s.
    2. If I had done the right steps thus far, I would have needed to subtract(?) 0.75, not multiply by 75.
    3. Trying again:

    Doing a unit conversion to m/s yields approximately 17.9(I rounded)
    Then I take my formula: v^2=vi+2aΔx
    Plugging in what I know: 0=17.9^2+2a(0.75)
    Simplifying: 0=320.41+2a(0.75)
    Subtracting 0.75: 0=319.66+2a
    Dividing by 2: a=159.83 m/s

    Is that right? Or should I have divided by 0.75?

    Thanks again for being so helpful!
     
  16. Oct 22, 2013 #15
    Oh, converting to gs: 16.3 gs.
     
  17. Oct 22, 2013 #16

    haruspex

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    That would have been better.
     
  18. Oct 22, 2013 #17
    Thanks! Trying one more time:
    Plugging in what I know: 0=17.9^2+2a(0.75)
    Simplifying: 0=320.41+2a(0.75)
    DIVIDING by 0.75: 0=427.22+2a
    Dividing by 2: a=213.61
    Dividing by 9.8 to convert to gs: 21.8 gs.

    Is that correct?

    Thank you again!!!
     
  19. Oct 22, 2013 #18

    haruspex

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    Looks good.
     
  20. Oct 22, 2013 #19
    Wonderful! Thank you!
     
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