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Homework Help: Please HelpInstantaneous Velocity

  1. Feb 18, 2004 #1
    I'm very confused...

    Please help me... It seems so simple..but its just not going into my head...

    Two cars travel in the same direction along a straight highway, one at a constant speed of 55 mi/h and the other at 70 mi/h.

    (a) Assuming that they start at the same point, how much sooner does the faster
    car arrive at a destination 10 mi away?
    (b) How far must the faster car travel before it has a 15 min lead on the slower

    How to solve a problem like this? Can Someone please Help?...

  2. jcsd
  3. Feb 18, 2004 #2

    Doc Al

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    Staff: Mentor

    Start by using Distance = Speed x Time, and see what you can figure out.

    For a) you know the distance, so find out how much time each takes.
  4. Feb 18, 2004 #3
    I got it.. but what about for the part ...
    How far must the faster car travel before it has a 15 min lead on the slower
    How do get that? Do I just solve distance=speed x time then get the dispalcement between carSlower and carFaster?

    So far this is what I got.
    .18h for the time of car running 55mph and .14h for car running 70mph from that I subtracted and got .04 and thats the answer I got for part 1 of my question.

    I solved for the distance for each car after 15 min.
    and got
    825mi for of car running 55mph and 1050mi for car running 70mph got the displacement and ended up with 225mi

    Is that right?

    There are no answers on the back of my book to check LOL
    Last edited: Feb 19, 2004
  5. Feb 19, 2004 #4

    Doc Al

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    Staff: Mentor

    Re: Re: Please Help..Instantaneous Velocity

    No. For part b you found how far each went in 15 minutes. That's NOT what you were asked to find. You saw in part a that after 10 miles, the fast car is ahead by 0.04 Hrs, so how far does it have to travel to be ahead by 15 min? That's the question.

    Let's review how to solve these problems.

    Let 1 be the fast car, 2 be the slow car. So:
    D = V1*T1
    D = V2*T2

    For part (a) we are given the distance and need to find the difference in time ΔT:
    ΔT = T2 - T1 = D/V2 - D/V1

    Which is what you did to solve a.

    Now for part (b) you are given ΔT (15min =.25 Hr) and have to find D. So take the equation above and solve for D.
    ΔT = D(1/V2 - 1/V1)

    You can plug in the numbers and solve for D. Or you can realize that...Ah... all that stuff to the right of D is a constant. So, ΔT is just proportional to D. So... you do the rest.
  6. Feb 19, 2004 #5
    THANKS.. I came up with the answer -3.75mi..

    Is that right?
  7. Feb 26, 2004 #6

    Doc Al

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    Staff: Mentor

    Not even close. The distance is negative??
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