Question

Consider two trains moving in opposite directions on the same track. The trains start simul­ taneously from two towns, Aville and Bville, separated by a distance d. Each train travels toward each other with constant speed v. A bee is initially located in front of the train in Aville. As the train departs Aville, the bee travels with speed u > v along the track towards Bville. When it encounters the second train, it instantaneously reverses direction until it encounters the first train, then it reverses again, etc. The bee continues flying between the two trains until it is crushed between the trains impacting each other. The purpose of this problem is to compute the total distance flown by the bee until it is crushed. Assume that the bee is faster than the trains. (a) Find an expression for the distance dn covered by the bee after its nth encounter with a train. Define d0 as the distance traveled during the first flight from Aville towards the train near Bville, d1 the distance traveled by the bee during the first trip from the Bville train to the Aville train, etc. Sum the resulting series to get the final answer. (b) Devise another way to obtain the same answer using very little calculation.

Answer 1

a)

1) event one:

do= ud[1/(u+v)]

2) event two:

d1= ud[(u-v)/(u+v)2]

3) event three:

d2= ud[(u-v)2/(u+v)3]

then:

s=do+d1+d2+...

= ud[1/(u+v)] + ud[(u-v)/(u+v)2] + ud[(u-v)2/(u+v)3] +...

s= ud{[1/(u+v)] + [(u-v)/(u+v)2] + [(u-v)2/(u+v)3] + ...}

answer: s=ud{d/(2v)}

b) Meeting time between trains.

T= d/(V+V)=d/(2V)

distance traveled by the bee:

s= T(u)

s= d/(2V)(u)

answer: s= ud/(2v)