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Running and Walking

  1. Jan 29, 2005 #1
    Tim and Rick both can run at speed v_r and walk at speed v_w, with v_r greater than v_w. They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.

    the time it takes tim to cover the distance is:
    t_R =((D/2)/v_w)+((D/2)/v_r)

    the time it takes rick to cover the distance is:
    t_T =2*D/(v_w+v_r)

    In terms of given quantities, by what amount of time, Delta t, does Tim beat Rick?
    It will help you check your answer if you simplify it algebraically and check the special case v_r = v_w.
    Express the difference in time, Delta t in terms of v_r, v_w, and D.

    this is an easy problem because all i need to do is take the Rick's time minus Tim's time right? i've done it and my answer is incorrect.
     
    Last edited: Jan 29, 2005
  2. jcsd
  3. Jan 29, 2005 #2
    You would take tim's time minus rick's time since tim takes less time. After that it is just algebra. All you have to do is simplify it.
     
  4. Jan 29, 2005 #3

    that's my problem. i couldn't simplify it correctly. i simplified it to be

    (2D(VwVr)-D/2(Vr+Vw)(Vw+Vr))/((Vw+Vr)(VwVr))
    but this is not the answer
     
  5. Oct 5, 2011 #4
    A. use the formula d=v/t and rearrange it for time to t=d/v, since you have two v values (v_r and v_w), change the equation to t= (d/2vw)+(d/2vr). The two is in there because you multiple the distance by (1/2) which puts the two into the denominator.
    B. The average speed is found by rearranging the answer to the above equation to answer for vw+vr, instead of time. You dont have to separate vw+vr when you get it on to the left side.
    E. you DO have to subtract tims time from ricks because even though tim finishes first, rick takes LONGER thus a greater value for time. (or time would be negative otherwise)
     
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