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Relative Motion Problems

  1. Sep 20, 2005 #1


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    I spend 2 hours this evening trying to do these problems, but for these 2, I couldnt manage to even draw a proper diagram for :frown: ... For the last question, I tried to draw it, but the resultant vector wasn't matching with the answer, so obviously it was wrong...
    Last edited: Sep 20, 2005
  2. jcsd
  3. Sep 20, 2005 #2


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    Let ve be the speed of the escalator, vw be your walking speed. Let the length of the escalator be x. You know rate x time = distance, so

    60ve = x


    90vw = x

    Therefore, 60ve = 90vw

    vw = (2/3)ve

    Can you work it from there?
  4. Sep 20, 2005 #3


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    I tried to work it out but doesnt that just restate the fact that walking is 2/3 times slower than escalator? Which you've already concluded from the 60 sec to 90 sec ratio...
  5. Sep 21, 2005 #4


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    This is really just like those "joint work" problems ("Billy can do a job in 2 hours, Rita can do the same job in 3 hours, how long would it take them working together?")

    The persons speed, walking only, is
    Riding the escalator without walking, the speed is

    Walking with the escalator going, the speeds add. The person's speed is
    [tex]\frac{1}{90}+ \frac{1}{60}= \frac{5}{180}\frac{"escalator"}{second}[/tex]

    The time to go up the escalator ("seconds per escalator") is the reciprocal of that:
    180/5= 36 seconds.

    First, draw a picture. The desired flight is "300 km [NE]" and she wants to fly that in 45 min. Okay, here "net" velocity vector must be 300/(3/4)= 400 km/hr (45 min is 3/4 hr). Draw a line at 45 degrees to the horizontal and to the right (NE) with "length" 400 km/hr. The wind is blowing from the north at 80 km/hr so draw a line up from your first line, with "length" 80 km/hr. Finally, draw the third line of the triangle- that will be the "heading" vector of the airplane. If you like you can draw "arrow heads" showing the direction along those lines. Notice that the arrow head on the "wind" vector is at the bottom. You want the first vector you drew to be the sum of the other two. That's why you drew the wind vector "from the north" upward.

    As always, there are two ways to do vector problems like this. One would be to break everything into coordinates. The coordinates for the wind are easy: 0i- 80j (i is to the east and j to the north, of course). The coordinates for the "true path" require trig: 400 cos(45)i+ 400 sin(45)j. Call the "heading" vector xi+ yj and add:
    xi+ (y-80)j= 400 cos(45)i+ 400 sin(45)j. That gives you two equations (treat the i and j components separately) to solve for x and y.

    However, since you give vectors by length and heading and give the answer that way, I suspect that you are intended to solve this by "solving the triangle". You have a triangle with two given sides of lengths 80 and 400. If you look closely at the triangle, you will see that the angle between those sides is 90+45= 135 degrees.
    You can use the cosine law to solve for the length of the third side (the speed the airplane should make through the air) and then the sine law to find the angle inside the angle, at the origin. Adding that to the 45 degrees that first side is from the horizontal gives the airplanes heading (N of E).
  6. Sep 21, 2005 #5


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    Ok I got it, thanks.
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