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Homework Help: Finding distance using acceleration and time

  1. Sep 22, 2010 #1
    Okay so this is the question i'm stuck on and dont know how to start:

    A car starts from rest at a stop sign. It accelerates at 4.2 for 6.0 , coasts for 2.3 , and then slows down at a rate of 3.3 for the next stop sign.

    please help!! thanks!!
  2. jcsd
  3. Sep 22, 2010 #2
    Do we get any units for this problem? Or shall I assume them to be m/s^2 and seconds?
  4. Sep 22, 2010 #3
    oohh sorry,,

    A car starts from rest at a stop sign. It accelerates at 4.2m/s^2 for 6.0s , coasts for 2.3s, and then slows down at a rate of 3.3m/s^2 for the next stop sign.

    How far apart are the stop signs???? find distance
  5. Sep 22, 2010 #4
    You need to use your equations of motion (suvat equations).

    s = ut + 0.5at^2

    where s = displacement, u = initial velocity, a = acceleration and t = time.

    You can use all of them for different types of questions like this so it is best to know them all, but for this one, you can use just the above one for the first two stages, and you will need another for the final stage. If you like, post your answer once you get it an I will happily check it for you.

    http://en.wikipedia.org/wiki/Equations_of_motion - You need the classic versions.
  6. Sep 22, 2010 #5
    yes i know that i'm supposed to use that eq'n, but what values do i put in them is what i'm confused about!
  7. Sep 22, 2010 #6
    k so for the first stage,

    D=0.5 (4.2)(6.0)^2
    = 75.6m

    Where do i go from there??
  8. Sep 22, 2010 #7
    Well let's take the first bit.

    You know a, u and t. So plug in the values and you get a value for s.

    EDIT: Above post noted.

    Right, know take the second bit.

    It coasts = no change in speed - acceleration = 0.

    Again, you now know a, u and t.

    EDIT: for u use v = u + at with the values from the first section. The v from this then becomes u for the second stage.
  9. Sep 22, 2010 #8
    i only know the initial velocity from where it started = 0m/s, and whats bothering me the most is that the question didn't say how long (time) it took the car to decelerate at the end?? so ur saying the first stage is right, now for the second and third stage... i think i will need more explanation please! please

  10. Sep 22, 2010 #9
    We don't need the time for the last bit. All in good time.

    The first stage is done.

    Now the second stage, do as I posted above.

    Remember, the final speed of the first section is the initial speed of the second section.
  11. Sep 22, 2010 #10
    uuum :S okay soo second stage...

    Vfinal = d/t
    = 75.6m/6 s
    =12.6 m/s

    is that right? now what? i have the final velocity of the first stage and initial velocity of the second stage///
  12. Sep 22, 2010 #11
    No, use the equation I gave: v = u + at.

    Where u = 0m/s, a = 4.2m/s^2 and t = 6.0. Plug those in and you should get your final speed for the first section. That is the initial speed for the second.
  13. Sep 22, 2010 #12
    v=25.2 m/s , now how can i use this velocity to determine the distance between two stop signs?
  14. Sep 22, 2010 #13
    A quick tip, if you write all the equations of motion out, along with their values (s,u,v,a,t), then for each stage, fill in the values you know.
    Once you have done this, look at the equations, find one which you have all but one value for and then you can plug in what you know to find the missing value.

    So for the first question you would have:

    from that you can see that you can use v = u + at to give the final speed and s = 0.5 ( u + v ) t to give the distance from point 2 to 3.

    That's how I do it.
    Last edited: Sep 22, 2010
  15. Sep 22, 2010 #14
    Right, for the second stage you now know u, v, a and t.

    So you need an equation all of these can go in, and the only thing you don't know is s.

    So in this case: s = 0.5 ( u + v ) t
  16. Sep 22, 2010 #15
    will this be it..?

    s = 0.5 (25.2 - 0)(8.3)
    ****8.3 = sum of all time
  17. Sep 22, 2010 #16
    right, because the car is coasting, the speed is considered constant (at least it is in this case). u = v

    the time is the time spent within stage 2 only. stage one is done, you don't need that anymore, forget that, you have the distance and final velocity which is all you need for the rest.
  18. Sep 22, 2010 #17
    okay so its 104.58m from stage one to the end of stage two? and now, in now how do i use the velocity and total distance to continue on to the third stage?

    how do incorporate the deceleration into it?
  19. Sep 22, 2010 #18
    hmm, I get a different distance for section 2.

    look at it as simply as possible:

    Distance = Speed x Time

    Distance = 25.2m/s x 2.3s = ? (hint: it aint 107)

    For the final stage you know u = (v from stage 2), v and a.

    Which equation of motion would you use? Remember, you don't need t, you just need s.
  20. Sep 22, 2010 #19
    v^2 = u^2 + 2a (Sfinal - Sinitial)?????
  21. Sep 22, 2010 #20
    Back on stage two for a second. You got it wrong.
  22. Sep 22, 2010 #21
    arrgh i'm confused and completely lost :( one post explaining everything from scratch with a final answer would be very nice :P
  23. Sep 22, 2010 #22
    OK, I'll do that

    All except the answer, that's down to you.
  24. Sep 22, 2010 #23
    thanks for all ur help!! will be waiting for that explanation :D
  25. Sep 22, 2010 #24


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    Homework Helper

    In the problem time for first and second stages are given. The time t3 for the third stage can be found by using
    vf = vi - at3 where vi = 25.2 m (which you have already found) and vf = 0.

    Now if you consider the time velocity graph, it is a trapezium. The total distance traveled is equal to the area of the trapezium.
    Area= d = 1/2*sum of the parallel sides* height.
    d = 1/2*v*(t1+t2+t3+t2)
    Substitute the values and find d.
  26. Sep 22, 2010 #25
    For the first section you know a, u, and t.

    Using s = ut + 0.5at^2 calculate the distance s for the first section.

    You also need v to use for the next two sections so work that out at the same time.

    Using v = u + at calculate the final speed v for section one.

    For section two, you know the initial speed is the final speed of section one. And because it is at constant speed initial speed = final speed (u = v). (No change in speed means acceleration = 0).

    So you know u = v, a and t.

    You can use s = ut + 0.5at^2 again to calculate the distance s for the second section.

    Now for the third section, u for section three = v for section two.

    You know u, v, and a.

    Using v^2 = u^2 + 2as you can calculate distance s. You need to rearrange this equation to give you s = ???

    Add up the three s values for the total distance.
    Last edited: Sep 23, 2010
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