I guess a better way to put it, it needs clarification, if a bird is sitting on a seat say and the train moves off, then starts flying, what happens? Nope Nothing like that - you'll kick yourself when someone tells you in a bit. These are all quite real questions, no trolling! Bleurgh beaten to it! OK the balloon one, when the car accelerates (fast) the air mass moves to the back of the car, and the pressure difference is enough to "float" the balloon forwards and hit the windscreen. Oo actually re-read yours - the wheel is correct but not quite. The *flange* of the wheel is always moving backwards, nothing too complex Here's another question: If you want to fly exactly to the opposite side of the world, what's the fastest route? (over the north / south, or fly east / west) ps i just made that one up, whilst still in bed so forgive if it's meaningless!
Assuming the Earth is a sphere it's not too complex. Include weather effects and the differing shape of the Earth it gets a bit more complex. The balloon in car assumes quite complex fluid movements of the internal air, if you were being kind you would discount them. The bird scenario is complex - if the train accelerates whilst the bird is connected they will move together, once the bird takes off it's down to the acceleration of the pair as to what happens, but the bird is always moving within the air on the train, rather than being completely independent of the train.
The question is very similar to the classical metaphysics/philosophical paradox, 'Achilles and the tortoise'. Classical philosophers would argue that the officer never reaches the front of the line in the first place. From Newton's time onwards we may apply calculus to solve this, and rightly so, the classical proof is obviously absurd. I have time to do neither.
Either east to west or west to east would prove to be fastest depending on the latitude you're at as aircraft will use jet streams to dramatically reduce flight time.
I still haven't seen an answer to the first question other than 66,66 To the trains question...regarding or disregarding slip?
This has been years since I've done that...I come out at..."they walk 125 meters" There's a catch somewhere, as with most school physics questions...always have to look for the catch.
For the first one it's just 200m, isn't it? The way I see it, on the trip to the front he travels 100m + the distance the troop travelled, on the return he travels 100m - the troop covered distance. If they're marching steadily it's just 200m total difference for the officer to travel to the front and back. That's how I've worked it on a scrap of paper anyway.
It's asking how much has the troupe of men traveled by the time the officer gets back to the start. They've been traveling three times slower = 3 times less distance. Initially I ignored the extra distance the officer had to travel for the first half, I just treated it like a typical drag race I live my life a quarter mile at a time.
Erm...surely it'd be 58.33m? You haven't taken into account that the troupe would be moving in the opposite direction to the officer. Your logic stands up on the journey to the head of the troupe, but not for the return as their relative velocities will be different each time. ...Did I just correct Nexxo about something? I'm actually wondering if I'm talking crap. EDIT: Wait no, I'm a moron. I think. I don't know. EDIT 2: Not writing this out was probably a bad idea.
I did read it, just my head glossed over the fact that it balances out overall. Moral of the story; don't question Nexxo's logic.
Argh! I also misread... i thought he moves forward, back and forward again... So only front to back. 75m then
ooh, a second way of calculating it. (in the first i calculated the two meeting points.) Relative speed... back to front his speed to the column is a factor two, so it takes hin 1/2 the time to reach the column head as it takes them to march... 1/2x100m=50m front to back the relative speed is 3+1, so 100m/4=25meters... Again. 75m. Wrong?
Hate this stupid question so much. Your solution is elegant, cancel out the unknown distance travelled (call it x) hence: (100+x)-(100-x)=200 Massive assumption, distance x is the same on the outward and return journey, this is incorrect, the officers relative speed to the troupe increases on the return journey (in this case doubles) so the distance travelled by the troupe is different each time. Solution: draw a graph with 3 elements, the officer, the troupe van and the troupe rear. The troupe van and rear are linear with a gradient 1, the troupe van starts at 100, the rear starts at 0. The officer's line has gradient 3 and starts at 0 too, at the point where it intersects with the troupe van line the gradient changes to -3. Finally the point where it intersects with the troupe rear line is the solution = 75m. Go away.
