2007-12-27 17:00:25 · 9 answers · asked by Que 1 in Science & Mathematics ➔ Mathematics
Require to calculate when the BACK of the convoy clears the tunnel.
Therefore, require the length of the convoy.
Let length of convoy be x yds
Back of convoy has to travel 1760 + x yd to clear tunnel.
Speed = 1760 yd / min
time = (1760 + x) / (1760) min
time = 1 + ( x / 1760 ) min
2007-12-30 05:49:39 · answer #1 · answered by Como 7 · 1⤊ 0⤋
a million. the two the practice and the tunnel are a million mile long, so in different words, how long does it take the practice to holiday 2 miles? distance travelled = velocity * time taken t = d / s = 2 / 30 = 0.06666... hours = 4 mins The practice exits the tunnel at 4.04pm 2. practice takes a minute and a a million/2 to holiday 3 miles. a million.5 min = 3/a hundred and twenty hours s = d / t = 3 / (3/a hundred and twenty) = a hundred and twenty mph
2016-12-18 10:01:49 · answer #2 · answered by ? 4 · 0⤊ 0⤋
It depends on how long the convoy was. It will take 1 minute, plus one minute for each mile of the convoy. If the convoy was half a mile long, then it will take 1.5 minutes.
2007-12-27 17:09:25 · answer #3 · answered by WhatWasThatNameAgain? 5 · 4⤊ 1⤋
If it kept travelling at the same rate it took 1 minute for
each person to go through.
Maybe they got spooked and didn't enter the tunnel.
How long is the convoy? Did they all make it through?
2007-12-27 17:08:37 · answer #4 · answered by pico t 2 · 1⤊ 0⤋
How long is the convoy?
You leave out a critical piece of information, Smarta**.
2007-12-27 17:09:55 · answer #5 · answered by Anonymous · 1⤊ 0⤋
it will be at least 1 min because you never said the convoy was traveling straight?
2007-12-27 17:06:42 · answer #6 · answered by Anonymous · 1⤊ 0⤋
It could take forever as you never classifed how long the convoy was
2007-12-27 17:56:39 · answer #7 · answered by Anonymous · 0⤊ 0⤋
60 seconds (lol)
2007-12-27 17:08:35 · answer #8 · answered by Anonymous · 1⤊ 0⤋
You've got to be kidding...
2007-12-27 17:06:35 · answer #9 · answered by endo_krono 2 · 1⤊ 0⤋