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Homework Help: Relative Velocity Questions

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data

    1.How long, in seconds, does it take an automobile traveling 50.0 km/h to become even with a car that is traveling in another lane at 40.0 km/h if the cars' front bumpers are initially 124 m apart?

    2.A shopper in a department store can walk up a stationary (stalled) escalator in 30.0 s. If the normally functioning escalator can carry the standing shopper to the next floor in 20.0 s, how long would it take the shopper to walk up the moving escalator? Assume the same walking effort for the shopper whether the escalator is stalled or moving.
    2. Relevant equations

    1. I assume, nothing besides conversion equations.

    2. Nothing really

    3. The attempt at a solution

    1. I immediately changed 124 m to .124 km. Then I tried to find out how long it would take the 50 km/h car to move .124 km farther. Then I realized that the distance between them still wouldn't be equal as the slower car would have still moved. How does this work? Is there something I'm missing?

    2. I tried subtracting the smaller time she took (20 seconds) from the longer time (30) to get 10 seconds but that was way too easy and of course didn't work. Then I tried to find how much longer is 30 seconds than 20 seconds. The answer 1.5, I then divided 20 by 1.5 to see the rate at which she walked but it seemed off so I'm lost.
  2. jcsd
  3. Oct 15, 2007 #2
    1) If you want to use relative velocity, which is a good way to do it, you can do it a couple of ways. You could look at the frame where the 50km/h car is at rest, and the car would see the other car coming at it at -10km/h. Or you could look at the frame where the 40km/h car is at rest, and it will look like the other car is coming at it at 10km/h.
    Alternatively, you can simply come up with a couple of equations and solve for them. The relative velocities merely cut one out, since they are equations themselves. Make sense?

    2) In the first case all you have is the shopper walking up the escalator. In the second case you have the velocity of the shopper in addition to the velocity of the escalator. There are two things that are held fixed in the problem, the distance to travel, and the velocity that shopper moves with. In the shopper on the broken escalator it is [itex]d = v_{\textrm{shopper}} * 30s[/itex], what would the other one be (the shopper on the escalator)?
  4. Oct 15, 2007 #3
    1. Yes, now I understand d= t x s so .124 = -10 x something. or .0124 hours or 44.64 seconds. Cool thanks

    2. The other one would be Vshopper = d x 20 seconds correct? But if you did that the velocity of the shopper is 0 (she is not moving) so distance has to equal zero which is not possible.
  5. Oct 15, 2007 #4
    The other would would be [itex]d = (v_s + v_E) *20[/itex]. Do you see why?
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