Question
Two points P and Q are connected by a straight road of length d. At time t = 0, a
car starts driving from P to Q at a constant speed c. At t = 0, a bee starts flying
from Q toward the car at a constant speed b > c, and takes a zig-zagging path,
reversing direction each time it meets the car or point Q. This process stops
when the car reaches Q. Let a1 be the distance the bee covers from Q to the car,
a2 be the distance it covers from the car back to Q, a3 be the distance it covers
from Q to meet the car the second time, and so on.
(a) Calculate the values of the infinite sequence a1, a2, a3, . . . of distances that
the bee covers.
(b) What is the total distance D = ai covered by the bee? Why should this
answer be expected?
[email protected]
car starts driving from P to Q at a constant speed c. At t = 0, a bee starts flying
from Q toward the car at a constant speed b > c, and takes a zig-zagging path,
reversing direction each time it meets the car or point Q. This process stops
when the car reaches Q. Let a1 be the distance the bee covers from Q to the car,
a2 be the distance it covers from the car back to Q, a3 be the distance it covers
from Q to meet the car the second time, and so on.
(a) Calculate the values of the infinite sequence a1, a2, a3, . . . of distances that
the bee covers.
(b) What is the total distance D = ai covered by the bee? Why should this
answer be expected?
[email protected]
Answers
As a sequence, let Tn be the time taken to meet the car for the nth time.
T1 = d/(b(c+1))
T2 = T1*b/(b(c+1))
...
Tn = d/(b(c+1))^n
The sum of this series, with
a = d/(b(c+1))
r = 1/((c+1))
is
S = a/(1-r)
= d/(b(c+1)) / (1 - 1/((c+1))
= d/bc
That is the total flying time. So, in that time, the be covers the distance d/c.
This is to be expected, since the car is approaching Q at speed c, the time taken is d/c.
The bee covers the distance bd/c in that time.
T1 = d/(b(c+1))
T2 = T1*b/(b(c+1))
...
Tn = d/(b(c+1))^n
The sum of this series, with
a = d/(b(c+1))
r = 1/((c+1))
is
S = a/(1-r)
= d/(b(c+1)) / (1 - 1/((c+1))
= d/bc
That is the total flying time. So, in that time, the be covers the distance d/c.
This is to be expected, since the car is approaching Q at speed c, the time taken is d/c.
The bee covers the distance bd/c in that time.
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