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Homework Help: Little help

  1. Sep 10, 2005 #1
    Not asking for an answer, just point me in the right direction please:

    In order to qualify for a racing event, a race car must achieve an average speed of 278km/h on a track with a total length of 1260m.

    If a particular car covers the first half of the track at an average speed of 209km/h, what minimum average speed must it have in the second half of the event in order to qualify? Answer in unts of km/h.

    So, I assume I have to split this up into two parts. The first half of the track he covers a distance of 630m at 209km/h or 58.05m/s. From there, you can derive a time of 10.85s (is time even needed?). But where do I go from here? For the second half of the track you you only know the distance. Which equation would I use to find missing variables?

    Also, could you explain why 278 = (209+v)/2 wouldn't work?

    Thank you.
  2. jcsd
  3. Sep 10, 2005 #2
    Let d = total distance, v_ave = average velocity over the entire track, d1 = distance of first part of track, d2 = distance of second part of track, v1 = velocity taken over d1, v2 = velocity taken over d2, t1 = time to drive over d1, t2 = time to drive over d2, and t_total = time to drive over d.

    Start with distance = velocity * time:
    d = v_ave * t_total, d1 = v1*t1, d2 = v2*t2.

    Since d = d1 + d2, we have

    v_ave*t_total = v1*t1 + v2*t2.

    We know v_ave and d, so we can find t_total. We know v1 and d1, so we can find t1 (as you did). We're trying to find v2, which can be found by first finding d2 and t2 and then dividing d2 by t2.

    To answer your second question, about why v_ave isn't equal to (v1+v2)/2, first take

    v_ave*t_total = v1*t1 + v2*t2,

    and solve for v_ave,

    v_ave = (v1*t1 + v2*t2) / t_total.

    By saying v_ave = (v1+v2)/2, you're implying that both t1/t_total and t2/t_total equal 1/2, which is untrue.
  4. Sep 11, 2005 #3


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    Homework Helper

    No, JoAuSc,
    OneTroubledGuy did NOT recognize v_ave ,
    but was ignoring it entirely.
    He's probably not yet in the habit of labelling
    his diagram with distances and times.
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