(A)A runner is jogging at a steady 7.7 km/hr. when the runner is 3.9 km. form the finish line, a bird begins flying from the runner to the finish line at 46.2 km/hr (6 times as fast as the runner). When the bird reaches the finish line, it turns around and flies back to the runner. How far does the bird travel? Even though the bird is a dod, assume that it occupies only one point in space and that it can turn without loss of speed.
(B) After this first encounter, the bird then turns around and flies from the runner back to the finish line, turns around again and flies back to the runner. the bird repeats the back and forth trips until the runner reaches the finish line. how far does the bird travel from the beginning ( including the distance traveled to the first encounter).
To solve this problem, we need to break it down into different parts and calculate the distances traveled by the bird in each part.
(A)
The runner's speed: 7.7 km/hr
The bird's speed: 46.2 km/hr
First, we need to calculate the time it takes for the runner to reach the finish line:
Distance = Speed × Time
Time = Distance / Speed
Time = 3.9 km / 7.7 km/hr = 0.506 hours
While the runner runs for 0.506 hours, the bird starts flying from the runner to the finish line.
Distance traveled by the bird from the runner to the finish line:
Distance = Speed × Time
Distance = 46.2 km/hr × 0.506 hours = 23.39 km
Since the bird reaches the finish line before the runner, it turns around and starts flying back towards the runner. The runner continues jogging towards the finish line.
The distance traveled by the bird from the finish line back to the runner will also be 23.39 km because it is the same distance traveled in opposite directions.
So, the total distance traveled by the bird in part A is:
23.39 km + 23.39 km = 46.78 km
(B)
In this part, the bird repeats the back and forth trips until the runner reaches the finish line. The runner is still jogging at a speed of 7.7 km/hr.
Since the bird travels the same distance from the runner to the finish line and back in each trip, we can calculate the number of back and forth trips it makes before the runner reaches the finish line.
Distance from the runner to the finish line: 3.9 km
Distance traveled by the bird in each trip: 23.39 km
Number of back and forth trips = Distance from the runner to the finish line / Distance traveled by the bird in each trip
Number of back and forth trips = 3.9 km / 23.39 km = 0.167
The number of back and forth trips will be rounded down to the nearest whole number (as fractions of an attempt are not possible). So the bird makes a total of 0 back and forth trips.
Now, we need to calculate the distance traveled by the bird in these back and forth trips:
Distance traveled by the bird in back and forth trips = Number of back and forth trips × Distance traveled by the bird in each trip
Distance traveled by the bird in back and forth trips = 0 × 23.39 km = 0 km
Therefore, the total distance traveled by the bird from the beginning, including the distance traveled to the first encounter, is:
Distance traveled in part A (46.78 km) + Distance traveled in part B (0 km) = 46.78 km
Flight time of bird to finish line, FL, = 3.9/46.2 = .0844 hr.
Distance runner travels during this time period = .0844(7.7) = .65km.
Distance from runner to FL = 3.9 - .65 = 3.25km.
Net closing speed between runner and bird = 7.7 + 46.2 = 53.9Km/hr.
Time for runner and bird to meet = 3.25/53.9 = .060 hr.
Distance covered by runner = .060(7.7) = .462km.
Distance covered by bird to runner = .060(46.2) = 2.785km.
Total distance traveled by bird up to first runner-bird meeting = 3.9 + 2.785 = 6.685km.
Remaining distance of runner to FL = 3.9 - .65 - .462 = 2.788km.
Time of runner to FL = 2.788/7.7 = .362 hr.
During this time period, the bird flies back and forth between the runner and the FL at the constant speed of 46.2km/hr covering a total distance of 46.2(.362) = 16.728km.
Therefore, the total distance traveled by the bird from the start = 6.685 + 16.728 = 23.413km.
I hope I didn't slip a digit or two.
Your problem brings to mind a golden oldie from the field of recreational mathematics that you might find amusing.
Two trains 150 miles apart are traveling toward each other along the same track. The first train goes 60 miles per hour; the second train rushes along at 90 miles per hour. A fly is hovering just above the nose of the first train. It buzzes from the first train to the second train, turns around immediately, flies back to the first train, and turns around again. It goes on flying back and forth between the two trains until they collide. If the fly's speed is 120 miles per hour, how far will it travel?
We want to know the total distance that the fly covers, so let's use Distance = Rate * Time to solve the problem. We already know the fly's rate of flight. If we can find the time that the fly spends in the air, we can figure out how far it travels.
Ignore the fly for a minute, and concentrate on the trains. The first train is traveling at 60 miles/ hour and the second train is going 90 miles/ hour, so they are approaching each other at 60 miles/ hour + 90 miles/ hour = 150 miles/ hour. Now we know the rate at which the trains are closing in on each other and their distance apart (150 miles), so we can find the time until they crash:
Distance = Rate * Time
Time = Distance / Rate
= (150 miles) / (150 miles/ hour)
= 1 hour.The fly spends the same amount of time traveling as the trains. It goes 120 miles/ hour, so in the one hour the trains take to collide, the fly will go 120 miles.