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Homework Help: Uniform accelerated motion question

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

    A sprinter runs a race with constant acceleration throughout. During the race he passes four posts A, B, C, D in a straight line such that |AB| = |BC| = |CD| = 20m. If the sprinter takes 5 seconds to go from A to B and 3 seconds to go from B to C, find out how long, the the nearest tenth of a second, it takes him to run from C to D

    2. Relevant equations
    s = (v + u)/2 x t

    3. The attempt at a solution

    I have a way of doing it but it takes up almost 1 A4 page :( I did it this way and the answer was wrong, wasn't surprised cos there was loads of room for error

    I did s = (v + u)/2 x t
    substituted (v - at) instead of u (v = u + at, u = v-at)
    And got v1 in terms of a
    Then did s = (v + u)/2 x t
    Substitued (u+at) instead of v
    Got u2 in terms of a

    v1 = u2 so I got the acceleration
    I kept doing this to get the u and v for C to D and used the acceleration to calculate the time
    But it was wrong

    I know there's a shorter way of doing it
    Can someone help me? Final answer is 2.4s (nearest tenth of a second)
  2. jcsd
  3. Nov 1, 2012 #2


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    What did you get for the acceleration? I make it 2/3, so if you did not get that please show your working that far.
  4. Nov 1, 2012 #3
    2/3ms^2 is what I got for acceleration
    Can this be solved as simultaneous equations?
  5. Nov 2, 2012 #4


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    You seem to have done the hard part ok. Please show your working from there.
  6. Nov 2, 2012 #5
    In the previous part, I got v |AB| = 4 + 2.5a so I solved that and got speed at B is 17/3 m/s, that's the u for |BC| so

    v = u + at
    v = 17/3 + 2/3 (3)
    v = 17/3 + 2
    v = 23/3 = speed at C

    Now the last part, the time from C to D
    u = 23/3 m/s a = 2/3 ms^2 s = 20m

    I tried to solve this with different UVAST formulae but I keep getting stuck or the wrong answer, typically something like 37 seconds which is impossible
  7. Nov 2, 2012 #6


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    A key idea I would use is that when accelerating at a constant rate of acceleration, the average velocity for a time interval is the instantaneous velocity at the middle of that time interval.

    NOTE: The middle refers to the middle time-wise, not the middle position.

    You can calculate the average velocity for a couple of time intervals from the date given.
  8. Nov 2, 2012 #7
    But how would you use the average velocity in this question?

    I solved it, thanks for your help guys
    I just had to use the v^2 = u^2 + 2as formula to get the final velocity at D
    Then t = v-u/a and solve for t
  9. Nov 2, 2012 #8


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    I agree with that. You must have gone wrong solving the quadratic. Check the signs.
  10. Nov 2, 2012 #9
    I didn't arrive at a quadratic for this question, I simply filled that into v^2 = u^2 + 2as. I don't even know where I went wrong earlier, must've been the different fractions
  11. Nov 2, 2012 #10


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    For the first 5 seconds, the average speed was 4m/s [covered 20m

    So 2.5 seconds into that 5 second period, the speed was 4.0 m/s

    For the next 3 seconds, the average speed was 20/3.m/s [covered the next 20m in 3 seconds]

    So 1.5 seconds into that time interval speed was 20/3 m/s

    so in the 4 seconds between those two times, the speed inceased by 8/3 m/s so acceleration is 2/3 m/s/s

    Once you have the acceleration you can calculate just about anything you want - as you did - this is just an alternate way to calculate the acceleration.

    Note - although the description took 2 minutes to type out, the mental calculation involved takes about 10 seconds to do.

    Note: with this acceleration, the runner will enter that third 20m with a speed of 23/3 m/s since that last section begins 1.5 seconds after speed of 20/3 m/s was reached [middle of the second interval, or 5.5 seconds after the speed of 4 m/s was reached [middle of the first interval.

    then s = ut + 0.5 at2 gives

    20 = 23/3 t + 1/3 t2


    60 = 23t + t2 [just getting rid of the fractions]

    t = -23 ± √(529 + 240)

    t = etc
  12. Nov 2, 2012 #11
    Thanks mate :) yeah that would've been a lot easier, that's handy enough to remember

    Ohh now I get how you ended up with the quadratic. Once I had u, s and a, I just used v2 = u2 = 2 as and after that it was just a matter of filling in the first formula t = v - u/a (v = u + at). The quadratic might've been shorter but it's easier to go wrong with it
  13. Nov 2, 2012 #12


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    I much prefer to use v2 = u2 + 2as then follow up with t = v - u/a (v = u + at) myself.
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