1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Acceleration and displacement

  1. Jan 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A car accelerates uniformly from rest along a straight road and experiences negligible frictional forces. The car travels 15m in the second second of its journey. How far does it travel in the fifth second? (OPTIONS: 15m, 35m, 45m, 75m)

    2. Relevant equations
    3. The attempt at a solution


    This seems really sophomore but I'm not arriving at any of those options. What I did was to consider that 15m was covered in two seconds (though I'm not completely sure if that's what is said, or if it means to say that 15m was covered in that specific second). So 15 / 22 = 3.75 ms-2. Multiplying this by 52 gives 93.75m.
    Plan B, assume 15m was covered in the specific second. Using a = (v - u) / t, v = 15ms-1, u = 0 and t = 2, giving 7.5ms-2.
    7.5 * 52 = 187.5s.
    ?
     
  2. jcsd
  3. Jan 20, 2016 #2

    gneill

    User Avatar

    Staff: Mentor

    The second second of its journey covers the time period t = 1 s to t = 2 s. The fifth second covers the interval t = 4 s to t = 5 s.

    Don't forget the velocity that the car has when it begins each interval.
     
  4. Jan 20, 2016 #3
    I don't follow. What's the error in my working? The units work out.
     
  5. Jan 20, 2016 #4

    gneill

    User Avatar

    Staff: Mentor

    The car did not cover 15 m in two seconds. It covered 15 m during the second second. That is, from time t = 1s to time t = 2s.
     
  6. Jan 20, 2016 #5
    Right, so u =/= 0. I'm not sure how to find u then.
     
  7. Jan 20, 2016 #6

    gneill

    User Avatar

    Staff: Mentor

    Work symbolically. Assuming a constant acceleration a, what would be the velocity at the end of the first second?
     
  8. Jan 20, 2016 #7
    a = v1 - 0 / 1
    a = v1?
    ----
    a = 15 - v1
    a = v1
    2v1 = 15
    v1 = a = 7.5
    No option for 7.5m. 7.5 * 5 = 37.5m.
     
  9. Jan 20, 2016 #8

    gneill

    User Avatar

    Staff: Mentor

    Working without units that is fine. A marker may be picky about dropping units, so you might want to declare a variable to represent a time interval of one second. Say, Δt = 1 s. Then you could write:

    ##v1 = a Δt##

    So, v1 will be the speed of the car as it begins its second second of the journey. What equation of motion can you write to find the distance traveled over that one second interval?
     
    Last edited: Jan 20, 2016
  10. Jan 20, 2016 #9
    Hmm. How was what I wrote wrong in practice?
    s = 0.5 * a * 12.
    I don't understand. I'd need one more value.
     
  11. Jan 20, 2016 #10

    gneill

    User Avatar

    Staff: Mentor

    That's not the whole of the basic equation of motion. What happened to the initial velocity?
     
  12. Jan 20, 2016 #11
    It's 0, because I'm looking at the first second as you suggested, and the car starts at rest.
    If you were to look at the second second, you'd have s = v1 + 0.5v1 = 1.5v1, assuming that v1 = a.
    EDIT: Ah, s = 15. So v1 = 10m/s?
    Just a note, I didn't think the question would be this long-winded. It's a 1 mark and this seems beyond what we've been taught.
    How was my value for 7.5ms-2 wrong? Or was it right?
     
  13. Jan 20, 2016 #12

    gneill

    User Avatar

    Staff: Mentor

    It will seem simple once you've had the "aha!" moment.

    Perhaps I wasn't clear in post #8. What I meant was, what is the equation of motion for the second one second interval (the "second second" so to speak), given that you now know the velocity going into that second is ##v1 = a \Delta t## ?
     
  14. Jan 20, 2016 #13
    Maybe, but now I have to understand why all my other working was wrong. :P Why was the 7.5 wrong?

    I edited my post just before yours, is that progress?
     
  15. Jan 20, 2016 #14
    If a is the acceleration and t is the time, what is the equation for the total distance s up to time t (assuming it starts from rest)?
    In terms of a, what is the total distance covered in 2 sec?
    In terms of a, what is the total distance covered in 1 sec?
    In terms of a, what is the total distance covered in 5 sec?
    In terms of a, what is the total distance covered in 4 sec?
     
  16. Jan 20, 2016 #15
    s = 0.5 * a * t2? I dunno, lol

    haruspex, would greatly appreciate your help if you're reading.
     
    Last edited: Jan 20, 2016
  17. Jan 20, 2016 #16

    gneill

    User Avatar

    Staff: Mentor

    Without units or comments explaining what the steps are, it is difficult for me to follow what you are trying to do there. When you replace a velocity with "a" because it's really "a ⋅ 1s" without a note stating that, things get muddy. That's why I suggested using the variable ##\Delta t## to represent a time interval of 1 second. Either hang on to the units or use variables to carry them along.

    Let's try a slightly different approach that might avoid some of the syntax issues.

    Since the vehicle is starting from rest and the acceleration is constant, the distance traveled versus time is simply:

    ##d(t) = \frac{1}{2} a t^2##

    Now you are given a distance traveled during a specific time interval, namely the second second of the journey. So use the above equation to find the expressions for the distances at either end of that interval. The difference between them should be your given value of distance for that interval. The only variable should be the acceleration which you can solve for.
     
  18. Jan 20, 2016 #17
    How can you possibly expect to be able to solve a problem like this if you can't even correctly state this basic formula?? lol, lol, lol.
     
  19. Jan 20, 2016 #18
    Throwing a cheap blow over a technicality? It was bad enough that you were useless.

    The only variable would be acceleration, but I don't know the initial velocity between t = 1 and t = 2?
    15 = u + 0.5 * a (for 2nd second, t = 1 so neglected)
    s = 0.5 * a (for 1st second)
    I need to understand why I was wrong for both the 7.5ms-2 = a value and the 10ms-1 = v1 final velocity after 1st second for this to click. Here's what I was doing in bold:
     
  20. Jan 20, 2016 #19

    gneill

    User Avatar

    Staff: Mentor

    You won't need it with this method. You can write explicit expressions for the total distance traveled up to any given time. Write the expression for t = 1 second and the expression for t = 2 seconds. The difference is the distance covered during the interval from t = 1 second to t = 2 seconds.
     
  21. Jan 20, 2016 #20
    The distance s travelled after 1 second is s = 0.5 * a
    The distance s travelled after 2 seconds is s = 2u + 0.5 * a * 22
    2u + 0.5 * a * 22 - 0.5 * a = 2u + 1.5a = 15 (second distance - first distance)
    Is that what you meant?
    This is really too much for a 1 mark.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Acceleration and displacement
Loading...