Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Kinematics problem

  1. Apr 6, 2006 #1
    Two rugby players are running towards each other. They are 37m apart. If one is accelerating from rest at 0.5m/s^2, and the other was already moving at 3.1 m/s and maintains his speed,

    How long before they hit each other?

    (if someone can guide me step-by-step that'd be great)

    (sorry, first timer on forum, my calcuations are on post #4)
    Last edited: Apr 6, 2006
  2. jcsd
  3. Apr 6, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You need to show some working and / or thoughts. We will then guide you through the problem.

  4. Apr 6, 2006 #3


    User Avatar
    Gold Member

    Welcome! In this forum people would be more willing to help you if you show some work!
    HINT: you know that they were running for the same amount of time, the sum of their distances should be 37.
  5. Apr 6, 2006 #4
    Calculations / work that i've tried (and failed):
    d = 37m
    V1 = 0
    V2 = ?
    a = 0.5m/s^2
    t = 7.5s

    So I tried to plug it into d = v1t + 1/2at^2 formula to find T..

    37 = 0t + 1/2(0.5)(t)^2
    37 = 0.25(t)^2
    148 = t^2
    t = 12.16s (7.5s is the answer)

    Thinking my answer was correct, I then realized that the accelerating person did not have to travel the full 37m to hit the other person, so therefore the accelerating person must've travelled less than 37m before hitting the other person. I've also tried to do two equations, one for each rugby player, then plug them into each other, since the time for both must be the same before they hit each other.

    Let d1 = distance traveled by accelerating rugby player before coming in contact
    d2 = distance traveled by other rugby player before coming in contact
    Something like: D1 = 37 - D2 (and vice versa)

    but then if i do that:
    37 - D = 0t + 1/2(0.5)(t)^2, wouldn't I have two variables...T and D then i'd be in a deeper hole...

    I really need some help, wondering if there are extra algebraic expressions I'm missing.

    (Excuse me for my poor sentence structure also, I need english help)

    please and thank you.
    Last edited: Apr 6, 2006
  6. Apr 6, 2006 #5


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Speed can be defined as;

    [tex]v = \frac{ds}{dt}[/tex]

    Hence, for constant velocity we can say that the displacement (s) is;

    [tex]s = v\Delta t[/tex]

    So, when the two rugby players collide we can say that if [itex]s_{1}[/itex] is the player travelling at a constant velocity that;

    [tex] s_{1} + s_{2} = 37 \Rightarrow s_{2} = 37 - v\Delta t[/tex]

    Now, what kinematic equation defines displacement in terms of time?

  7. Apr 6, 2006 #6
    Sorry for my lack of understanding...

    I know the equations that define displacement in terms of time, they are:

    d = v1t + 1/2a(t)^2
    d = 1/2(v1 + v2)t

    The answer that i'm looking for is the amount of time it took (from start to finish) for the two rugby players to come in contact

    Now, the 37-vt you refer to...is the V referring to v1? or v2?

    Starting to get frustrating after doing the same question for an hour -_-

    Where do I plug in the 37-vt ?
  8. Apr 7, 2006 #7
    There are at least 2 ways of solving this and Hoots suggestion seems to be in the spirit of the tools given to you thus far
    Call the accellerating dude A, and the dude running at constant 3.1m/s B
    How would you define the distance (any distance) travelled for A...how would you define the distance travelled for B (again...any distance)..add both these expressions as Hoot suggested and you will know what they are equal too...then you need to solve :wink:

    Another way but I'm betting this tool hasn't been given to you just yet...you could consider the motion of A relative to B...ie; from B's point of view he is running at 3.1m/s but if he looks around him, everything is moving towards him 3.1m/s quicker than it would be if he was standing still...now lets imagine he think's he's on a treadmill and that everything *really is* travelling towards him at 3.1m/s . Can you see a trick that you can employ with this line of thinking?
    incidently...both approaches lead to the same equation
    Last edited: Apr 7, 2006
  9. Apr 7, 2006 #8


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    In the 37 - vt, v is the velocity of the non accelerating body. I.e. the one travelling at 3.1 m/s.

    You should use your first equation [itex]s_{2} = v_{1} + \frac{1}{2}at^2[/itex] (I've just changed the d to [itex]s_{s}[/itex] to fit my notation. As the accelerating player starts from rest the initial velocity term drops out leaving you with two equations;

    [tex]s_{2} = 37 - vt[/tex]

    [tex]s_{2} = \frac{1}{2}at^2[/tex]

    You now have two equations which equal [itex]s_{2}[/itex]. Therefore, you can equate them;

    [tex] 37 - vt = s_{2} = \frac{1}{2}at^2 \Rightarrow 37 - vt = \frac{1}{2}at^2[/tex]

    Do you follow? All that remains is to solve for t. Remember that v is the velocity of the non-accelerating body (3.1 m/s)

    As Greg A said, another method is to consider relative velocities.

    Last edited: Apr 7, 2006
  10. Apr 7, 2006 #9
    Thank you all, I've finally understood how to do the problem.

    I'll definitely be referring my friends to this forum.

    Thanks again.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook