6no one tryn??
A rubber band with initial length L has one end tied to a wall. At t = 0, the other end is pulled away from the wall at speed V (assume that the rubber band stretches uniformly). At the same time, an ant located at the end not attached to the wall begins to crawl toward the wall, with speed u relative to the band. Will the ant reach the wall? If so, how much time will it take?
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11 Answers
1My first answer will be nop... cuz in practical situations, ur comparing an ANT with a RUBBER BAND, hence ant cannot reach the wall before the rubber band does.
My second answer is that, if the ant and the rubber band move along the same line of action, the rubber band will push the ant along with itself. Since the ant is walking within the stretched area of rubber band, the rubber band will carry the ant to the wall. So the ant touches wall first, then the rubber band does.
My third answer is that, the ant will die, because the rubber band's force will be large enough to kill it!
My fourth answer is completely theoretically based on physics.
u is the velocity of any w.r.t rubber band. If u is +ve then ant reaches the wall, else it does not.
6ankur frankly speaking i hav'nt tried bt i felt like postingthis one.i'll try it later and give my ans.
1lol... u cnt hv an exact answer fr this qsn. cuz practically its nt possible, logically the ant dies, and by theory if physics, the last ans of mine has smthing to do with it [3]
6wat if the ant has legs of superman ....if it is antman???:D:P then wat??
49i feel answer is Lu-V2 seconds..
seems i ultimately solved this one after 4 months 15 days..[3][3][3]
49posting solution..
the rubber bands free end will move back with a velocity V its fixed end will move with velocity 0
moving from fixed end to free end, the velocity of each point on rubber band will increase linearly depending on the as a function of distace from free end...i can prove this
thus when the ant moves from free end to fixed end it will face a BACKWARD VELOCITY due to moving rubber band and this will be linearly decreasing..
so as an equivalent case we can take the average backward velocity V+02 to be faced by ant throughout the journey...
thus w.r.t ground the constant velocity can be taken to be u-V2
hence the answer!!
66At any instant t after the beginning, the instantaneous length of the string is L+Vt and so the velocity of a point at a distance x from the fixed end is xL+Vt V. So if the ant is at this x, its velocity w.r.t. to the wall is xL+Vt V &ndash u. Hence we obtain the differential equation
dxdt = xL+Vt V &ndash u
Its not difficult to solve this equation. Using the initial condition that at t=0, the ant was at a distance x0 from the wall, the instantaneous position x can be obtained as
x = x0 (1+VtL) &ndash uLV(1+VtL) ln (1+VtL)
The time taken to reach the wall can be obtained by setting x=0, which gives the required time as
t = LV(ex0V/Lu &ndash 1)
which is always finite (unless, of course, u =0; but this case is obvious).
6well yes logically the ant wont and cant reach the wall:D
but yes trying to solve the quesn out of it!!!