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Acceleration of a car given distance and time?

  1. Sep 16, 2012 #1
    This question was on my practice exam last week. However, I had no idea how to solve it, since plugging the variables into formulas left me with TWO unknown variables (no matter what formula), making any of them impossible to solve.. But there must be a solution. I'd be very grateful if you could give this a try?

    1. The problem statement, all variables and given/known data

    A car accelerates from rest for 4 seconds, then maintains a constant velocity for 14 seconds. At the end of 18 seconds, it has traveled 1200 m. What is
    a) its acceleration for the initial 4 seconds
    b) the distance at which it stopped accelerating
    now assume that the car decelerates at -12.5 m/s
    c) how long does it take for it to come to a full stop, and after how many meters?


    2. Relevant equations

    No idea, but I think one of:
    v = v(initial) + at
    v^2 = v(initial)^2 + 2a(x - x0)
    or other one-dimensional equations?

    3. The attempt at a solution

    I tried plugging the variables into every equation, but all of them yielded two unknown variables, making them impossible to solve.
    Since I knew that in the second phase (phase with constant velocity) the initial and final velocities were the same, I did try replacing them with zero, as if I'd subtracted one from the other.. But that didn't work.
     
    Last edited: Sep 16, 2012
  2. jcsd
  3. Sep 16, 2012 #2
    What is the distance travelled when it is in acceleration mode?
    To find the average acceleration, use SUVAT equation.
     
  4. Sep 16, 2012 #3

    Ibix

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    Unless the initial velocity of the car is given, I think you are correct. I rather suspect that the question should read "a car accelerates from rest for 4 seconds...", in which case it's soluble as per azizlwl's comment. But that's not something you should assume unless there's context to the question that you haven't given us.
     
  5. Sep 16, 2012 #4
    Oh, I'm sorry. I think I forgot that aspect of the problem... Yes, the car accelerates from rest. Also, I'm not sure how I should solve it even after reading azizlwl's comment..?
     
  6. Sep 16, 2012 #5

    Simon Bridge

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    It is very common to get confused when you try to think which equation to use ... you get mixed up. When that happens in kinematics, the best thing you can do it quickly sketch the v-t diagram.

    So draw it for the motion:

    at t=0, v=0
    at t=4s, v=? ... just put a mark on the v axis and label it v.
    Draw a point at (4,v) and connect it to (0,0) ... that is the first leg of the journey.

    The next step is a constant velocity for 14 seconds ... so the final times is t=4+14=18s and the final velocity is still v. So put a dot at (18,v) and join it up to the last dot with a horizontal line.

    Now there are only two things to remember about a v-t graph:
    the slope is the acceleration
    the area under the graph is the displacement

    So, for the 1st 4 seconds - the slope is v/4 = a1 and the displacement is (how do you find the area of a triangle) s1=0.5x4xv=2v

    That's all you can do.
    For the second part of the motion, it is just a rectangle ... you can find a2 and d2 yourself right?

    This isn't good enough to find the answers to (a) and (b) yet ... there's one more thing: you are told that the total displacement is 1200m = d1+d2

    You have equations for d1 and d2, so substitute them in there.
    Now you can find v and the rest will be easy.

    For the third part - go back to the v-t diagram, and and another bit on it.
    It is going to be a deceleration - you don't know the time yet so just put a mark on the t axis and call it T

    Do the same thing - the slope is rise over run = acceleration ... this time you are told the acceleration, you know v, you have to find T-18 (why not T?).
     
  7. Sep 16, 2012 #6
    SUVAT
    v(final) = v(initial) + at
    v^2 = v(initial)^2 + 2a(x - x0)
    x=x0+ ut+(1/2)at2
    ---------------------------------------------

    Data given are final velocity, time and distance(after substracting from whole journey).
    Take the acceleration as average and use 2 of the SUVAT equations to solve for a.
     
    Last edited: Sep 16, 2012
  8. Sep 16, 2012 #7

    Ibix

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    If you know that the car starts from rest, the problem is soluble. I'd start with s=ut+(1/2)at2 and apply it separately to the two parts of the car's motion.

    Edit: ...although there are other methods, as you see.
     
  9. Sep 16, 2012 #8

    Simon Bridge

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    It is good pedagogy: someone is getting confused - take them back to basics.
    Anyway - it is possible that someone having trouble picking an equation is a visual thinker ;)

    The trouble with the "pick an equation, any equation" approach is that, when it goes wrong, the overwhelming temptation is to just tell them the right equation to pick. But their problem is that they have not learned how to pick the right equation - better to tell them this metaskill.

    It's just that it is so boring ... list what you know, compare with the kinematic equations, pick the one that has all the stuff you do know and only one of the stuff you don't know, numbers in, stains out - it's like magic!

    Of course the trouble with the graphical method is that this is a text-intensive medium (and people seem scared of drawing pictures for some reason...) and there is something reassuring, isn't there, about knowing you have the right equation ready... somewhere...
     
  10. Sep 16, 2012 #9
    Ok, I think I've got it...

    a) 18.75m/s^2
    b) 150 m
    c) 6 seconds, 1425 m including the first 1200

    Is this right? And thank you everyone, for all the help!
     
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