Any math whizzs here?  Two bugs are on an east-west line a distance d apart. both travel at the same speed.---
826981
It looks to me like it's .5d which is not too far from your answer.\n\nFor simplicity, use a unit of length in which d = 1 and a unit of time in which the speed of each of the two bugs is also 1. Let r be the distance between them and let h be the angle between the y direction and the vector from S to R. In other words h starts out pi/2 and approaches 0.\n\nThe time derivatives are:\n\ndh/dt = -sin(h)/r , dr/dt = -1 + cos(h)\n--->\ndr/dh = r(1-cos(h))/sin(h)\n\nSeparating the variables r and h to opposite sides of the equation and integrating both sides with the initial condition (r=1 at h=PI/2) gives:\n\nr = tan(h/2)/sin(h)\n\nYou want the distance between the two bugs as t-->oo or equivalently as h --> 0 which is r=1/2 (use L'hopital's rule for example), or d/2 in terms of the initial distance d.\n\nOf course S never actually reaches the y axis but approaches it asymptotically.\n\n(Maybe I should add, I first tried to formulate the problem in terms of the Cartesian coordinates and was able to write a differential equation for the curve but it looked very difficult to solve. Finally tried using the above coordinates (r,h) and it became much easier)
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
164737
if it can travel at the same speed at a 45 degree angle, they will both travel the same distance.
711513
Hi!  You are right.\nIf they have equal masses and are elastic then  b) it is going to happen. Why? well Vf= 2*Vcm-Vi. Where all are vectors and :\nVf= final speed;   Vcm=center of mass speed ;  Vi=initial speed.\nThus 0=2*Vcm-Vi => Vcm=V/2=(m1V1+m2V2)/(m1+m2)=\n=m1*Vi/(m1+m2)  => m1=m2.    \n  " a)" isn't possible because the total momentum must be preserved.\n          Hope it helps!
530622
Ok, first you want to see where they meet. To make this easier, advance train A for 5/6ths of an hour and see the problem in terms of when train B starts out. 65 x 5/6= 325/6\nkilometers. So the problem is the same only they start out at the same time over a distance of 600- 325/6 kilometer =\n545 5/6 kilometers.\n\nNow, they travel in the proportion of 65/105 kilometers per hour. So the distance till they meet will be in the same proportion. That is 545 5/6 x 65/105 = 212875/630 or about 338 kilometers for train A to travel. Since it travels at 65 kph, this means it travels for 5.2 hours from the new start time equal to B's which was 6:20 AM. Ends up being 11:32AM.
53689
The combined speed of the two cyclists is 25k/hr\nThey will meet in 2 hours.\nThe bee is flying 20k/hr for 2 hrs so it will fly 40K.
1022304
Point A and Point B are 600 miles apart. A train leaves Point A travelling toward Point B at 5:30AM at a constant speed of 65 kph. Another train leaves Point B heading toward Point A at 6:20AM at a constant speed of 40 kph. What time do the two trains meet?1 kilometer = 0.6214 miles\n\nx=kilometers     5/6(50 minutes/60 minutes)*65= 54.166...km\nx=.6214 (600 miles)\nx=372.84 kilometers apart\n5/6(50 minutes/60 minutes)*65= 54.166...km\n372.84-54.17= 318.67km between the 2 trains at 6:20\nthats as far as I can get (I don't remember how to do this... hope this helps!)
44361
Wind direction/resistance would be a factor, but if we disregard the wind and the planes are running at the same airspeed, the one heading westward would theoretically have a shorter flight.  This is because the Earth's rotation is eastward.  So as the plane flies westward toward it's destination, that city is also moving eastward toward the plane.  The opposite is true for the plane heading eastward.  It's destination is also moving eastward away from the plane.  While there wouldn't be much difference, the westward plane should make the flight a little more quickly.
104070
Heather leaves a town at 2 p.m., going 56 mph, traveling east. Erin leaves the same town at the same time going 64 mph, traveling in the same direction. What time of day will it be when Erin is 72 miles ahead of Heather?\n============================================================\n\nHeather's distance = 56x\nErin's distance = 64x\nx = hours traveled\n\nThe distance between them at any time can be expressed as a simple equation:\n\nd = 64x - 56x\nd = 8x\n\nYou want to find "x" when d = 72 miles:\n\n8x = 72\nx = 9 hours\n\n2:00 p.m. + 9 hours = 11:00 p.m.
1060398
Two airplanes are both travelling along linear courses, at fixed speeds.  Both will pass over a specific place.  At the same time, one plane is 500 miles east of the place; the other plane is 300 miles south of the place.  If the east plane travels at 500 mph, what speed must the other plane travel at to collide with the first?\n\n--keep in mind I've created a 3D system here--\nSet the point of collision at the origin (0,0). Now we have:\nx=500-500t\ny=300-nt.\nWe're looking for n.\nWe know that t=t in order for them to collide. \nNow solve for x=0 to find out when (the value of t) the first plane reaches the place of collision.  t=1.\nPlace y=0 and t=1 to solve for n.\nn=300mph.\n\nYou have to make your own graph.
196187
Okay, you are talking about two points in time:  1PM and 3PM.  This implies that the beginning point is the same.  This means that the final points will indicate a triangular shape.  Since the directions are north and west, these are perpendicular directions, meaning the triangle formed would be a right triangle.\n\nThe first plane traveling north  will be able to travel 300 miles and the second plane will be able to travel 400 miles.\n\n(2 hours x 150 miles per hour = 300 miles)\n(2 hours x 200 miles per hour = 400 miles)\n\nSo, this means that these are the two legs of the right triangle.\n\nusing the pythagorean theorem, \n\na^2 + b^2 = c^2\n300^2 + 400^2 = c^2\n90000 + 160000 = c^2\n250000 = c^2\nc = 500 miles apart
413991
They are moving towards each other at an increasing rate.
1004891
Nope...force is what you are looking to determine.  Force equals mass x acceleration.  The fly would have to travel MUCH faster to have any impact on the train.  Also, the fly, once it made initial contact with the train, doesn't decelerate all at once...parts of it continue moving after other parts stop, and some parts start moving in East and West directions...SPLAT!
349797
There is a bug regarding this....
913200
Both their accelerations will be the same and equal to the acceleration due to gravity, g=9.81 m/s/s...This is coz the only force acting on them is the force due to gravity
732600
SRY!! Matt is going to a basketball game out of town. If he could travel 75mph, He would arrive one hour early. If he could travel 45mph he arrives one hour too Late. How far is Matt's trip\n\ndis=75*(time-1)\ndis=45*(time+1)\n\nthen\ndis= 225mile,    time=4hour
780136
Different trajectory?