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| Thread ID: 87276 | 2008-02-15 01:21:00 | maths equation | caffy (2665) | PC World Chat |
| Post ID | Timestamp | Content | User | ||
| 640498 | 2008-02-15 01:21:00 | Hi, This is a totally random post, but I'm curious :) While travelling around Europe last year, many of the airports we visited had moving walkways. One airport in particular had 2 walkways next to each other, and the speed of one was 3kph and the other was 9kph (there were signs telling you how fast it was). Now, if the average person walks, say, 4kph...how fast will that person be if they walk at 4kph on the 3kph moving walkway? And the 9kph walkway? Do I times the 4kph by 3kph to get 12kph? I last did maths in 2003 so I don't remember the equation for speed, and particularly when you want to add more variables into the equation! |
caffy (2665) | ||
| 640499 | 2008-02-15 01:29:00 | Wouldn't it simply be 4kmh (Walking Speed) + 3kmh (or 9kmh), so on the slower walkway, the person would be moving at 7kmh, and on the faster one, 13kmh. Relative to the ground in this case, which obviously isn't moving. |
wratterus (105) | ||
| 640500 | 2008-02-15 01:52:00 | I last did maths in 1962, and failed miserably, yet I still know that you just add the two speeds together. Your speed relative to the walkway is additive in relation to the speed of the walkway relative to the ground. Walk in the opposite direction and your speed relative to the ground will be 1km/hr. The equivalent values apply for your second example. However, based on previous experience with the airplane thread, you should be aware that some may say that you will stand still, or your feet will explode, or maybe even your legs will shear off at the knees. Cheers Billy 8-{) :D |
Billy T (70) | ||
| 640501 | 2008-02-15 03:27:00 | OR to be completely silly - and don't say you've never tried this :rolleyes: get to the end and walk back the wrong way - So if the walkway is moving at 3kph you would have to walking at least 4 kph to get any where - even then it would be at 1 kph - how long was that walk way :lol: OR on the 9kph you would have to doing at least 10kph + (is that a jog speed?) other wise you would be going nowhere fast :p and you couldn't stop either ,other wise you would be right back where you started at the end - or is that the beginning:confused: ?? |
wainuitech (129) | ||
| 640502 | 2008-02-15 04:17:00 | The way I see it as this: You're a swimmer swimming in a completely still pool (or beach), swimming 4kph, then suddenly the current picks up (3kph current) and you are now MOVING (key word) at 7kph. It doesn't matter about your feet, or timing or anything complicated like that, you are still MOVING at your speed + the environment speed. |
--Wolf-- (128) | ||
| 640503 | 2008-02-16 02:38:00 | cheers for that, it makes sense. And of course I have gone to the end of an escalator and tried walking down it! :) not sure if I did on moving walkways though |
caffy (2665) | ||
| 640504 | 2008-02-16 03:11:00 | However, based on previous experience with the airplane thread But what if the ground was moving in the opposite direction, at the same speed? |
wotz (335) | ||
| 640505 | 2008-02-16 06:05:00 | Wolfforest's tidal analogy is spot on the mark - the result is the relative speed in relation to the ground - not the medium be it tide or air. If the tide was at an angle to the swimmer then it would be resolved into a downtrack and cross track component. | KenESmith (6287) | ||
| 640506 | 2008-02-16 11:55:00 | It is important to understand that all velocities are relative. You are walking at 3km/h in the forward direction relative to the walkway and the walkway is moving at 4 km/h relative to the ground. Since both velocities are in the same direction we can simply add the speeds to get your velocity relative to the ground - 7 km/h in the forward direction. If you were walking in a different direction to the walkway it would be a little more complicated. en.wikipedia.org |
TGoddard (7263) | ||
| 640507 | 2008-02-16 23:46:00 | Wolfforest's tidal analogy is spot on the mark - the result is the relative speed in relation to the ground - not the medium be it tide or air. If the tide was at an angle to the swimmer then it would be resolved into a downtrack and cross track component. Really? Most of your post doesn't even comprehend in my brain, yet I made a spot on analogy? Yay for 5th form maths. |
--Wolf-- (128) | ||
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