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?
ramaldon
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.
To calculate the values of the infinite sequence a1, a2, a3, ..., we can break down the problem into smaller parts.
Step 1: Distance traveled while the car is driving towards Q
When the car starts, the bee starts at Q and flies towards the car. Since the car and bee are moving towards each other, their relative speed is b + c.
Since the distance between P and Q is d, and the relative speed of the car and bee is b + c, the time taken for the car to reach Q is given by t1 = d / (b + c).
During this time, the bee covers a distance a1 = b * t1.
Step 2: Distance traveled while the car is returning from Q to meet the bee
Once the car reaches Q, it turns around and starts driving towards the bee. The bee also changes direction and flies back towards Q.
Since the bee is now moving towards the car, their relative speed is b - c.
The car travels the same distance d again (from Q to P) at a speed c, so the time taken is t2 = d / c.
During this time, the bee covers a distance a2 = (b - c) * t2.
Step 3: Distance traveled while the car is driving towards Q again
Once again, when the car reaches the bee, it turns around and starts driving towards Q. The bee changes direction and flies towards the approaching car.
The time taken for the car to reach the bee is t3 = d / (b + c), the same as in Step 1.
The bee covers a distance a3 = b * t3.
This process continues indefinitely, so the distances covered by the bee will form an infinite sequence: a1, a2, a3, ...
To calculate the total distance D, we can sum up all the distances covered by the bee:
D = a1 + a2 + a3 + ...
To find a closed-form expression for the sum, we can observe that the odd-indexed terms (a1, a3, a5, ...) are constant multiples of t1, and the even-indexed terms (a2, a4, a6, ...) are constant multiples of t2.
Using this observation, we can express D in terms of t1 and t2:
D = (b * t1) + ((b - c) * t2) + (b * t1) + ((b - c) * t2) + ...
Since the terms a1 and a3 are the same, and a2 and a4 are the same, we can rewrite this as:
D = 2 * (b * t1) + 2 * ((b - c) * t2) + 2 * (b * t1) + 2 * ((b - c) * t2) + ...
Simplifying this further, we get:
D = 4 * (b * t1 + (b - c) * t2)
Finally, substituting the expressions for t1 and t2, we get:
D = 4 * ((b * d) / (b + c) + ((b - c) * d) / c)
So the total distance D covered by the bee is 4 * ((b * d) / (b + c) + ((b - c) * d) / c).
This answer is expected because as the car and bee move towards each other, their relative speed decreases. As a result, the bee covers smaller distances in each subsequent round trip. However, since there are infinitely many round trips, the total distance covered by the bee can still be significant.