Consider two bulldozers in a tunnel moving towards each other in a straight line
ID: 1533076 • Letter: C
Question
Consider two bulldozers in a tunnel moving towards each other in a straight line at a constant speed of k miles per hour on a collision course. At time t = 0, one bulldozer was located at position x = 0 and the other bulldozer was located at position x = d (that is. when the bulldozers began their motion they were separated by a distance of d miles). Suppose that at lime t = 0 a bee was perched on the front of the bulldozer at position x = 0. The bee. which has no chance of escaping, begins to fly back and forth between the bulldozers in a straight line at a constant speed of:v miles per hour, desperately trying to avoid its fate. Unfortunately, the bee is eventually crushed. Note that v > k. Let x_n denote the distance of the bee from its starting point (that is. the t = 0 position) when it makes its n^th landing on a bulldozer. Derive an expression for x_n. At what location relative to the initial location of either bulldozer will the bee be crushed?Explanation / Answer
(a) In first term the distance d is covered by fly and bulldozer in time t =d/(k+v) (time = distance/ speed )
so distance tavelled by fly in this time t is x1=(d/(k+v))v = dv /(k+v)
now in 2nd iteration
the the disatance covered by two bulldozer in time t=d/(k+v) is
d=(d/(k+v))2k
so remaining distance between bulldozer is =d-(2kd/(k+v))
=(dv-dk) / (k+v )
now this distance is covered by bulldozer and fly in time t2 =((dv-dk) / (k+v ) ) / (k+v)
= (dv-dk) / (k+v )2
and distance covered by fly in this t2 time is =((dv-dk) / (k+v )2 )v =(v(dv-dk) / (k+v )2 )
and distance covered by fly in this time is x2=dv /(k+v) - (dv(v-k) / (k+v )2 )
so xn=dv /(k+v) - (dv(v-k) / (k+v )2 ) + ...............+ (-1)n-1dv(v-k)n-2 (1/(k+v)n)
(b) the bee will be crushed at the position d/2 . Since the bulldozer travells with same speed and opposite direction so it will take place half way from starting point.