HELP! I don't understand this physics problem?
A runner is jogging in a straight line at a
steady vr= 2.2 km/hr. When the runner is
L= 7.1 km from the finish line, a bird begins
flying straight from the runner to the finish
line at vb= 4.4 km/hr (2 times as fast as the
runner). When the bird reaches the finish
line, it turns around and flies directly back to the runner. What is the cumulative distance the bird travels?
Part 2: 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
- 11 months agoFavorite Answer
Your question does not make it clear. Is this a once round trip for the bird, or a repeated back and forth until the runner arrives at the finish line?
Let's assume that the question is as stated. Bird flies from runner to finish line and then back to runner. End of story.
Follow the bird. How long will it take the bird to reach the finish line?
speed = distance/time
time = distance / speed = (7.1 km / 4.4 km/hr) = 1.61 hr
[freeze the picture]
So after 1.61 hours, we have a bird at the finish line that has flown 7.1 km and has turned around.
Meanwhile, the runner moving at half speed has only run (7.1 km/2) = 3.55 km, so he is still 3.55 km from the finish line.
Now we have a runner and a bird 3.55 km apart moving towards each other at a closing speed of
(2.2 km/hr + 4.4 km/hr) = 6.6 km/hr. How long until they meet?
time = distance / speed = (3.55 km / 6.6 km/hr) = 0.538 hr
How far has the bird flown (moving at 4.4 km/hr) in that 0.538 hr?
distance = speed * time = (4.4 km/hr * 0.538hr) = 2.367 km
Total distance flown by bird = 7.1 km + 2.367 km = 9.467 km
Say 9.5 km to two significant figures.
- Anonymous11 months ago
The bird travels twice as fast as the runner.
When they meet again, the runner has traveled 2/3 of the 7.1 km and the bird has traveled 4/3 (3/3 out and 1/3 back) of the 7.1 km, or 9.5 km. ◄
If the bird flew 4 times as fast, then when they meet again the runner would have traveled 2/5 of the 7.1 km and the bird would have traveled 8/5 of the 7.1 km.
If the bird flew 10 times as fast, then... runner = 2/11 and bird 20/11
I conclude that if the bird travels N times as fast as the runner, then when they meet again the runner has traveled 2/(N+1) of the distance and the bird has traveled 2N/(N+1)
Go back to our case -- N = 2
so runner distance = 2/(2+1) = 2/3 (of the initial distance to finish line)
and bird distance = 2*2/(2+1) = 4/3 (of the initial distance to finish line)
Hope this helps!
- oubaasLv 711 months ago
A runner is jogging in a straight line at a steady Vr = 2.2 km/hr.
When the runner is Lr = 7.1 km from the finish line, a bird begins flying straight from the runner to the finish line at Vb = 4.4 km/hr (2 times as fast as the
runner). When the bird reaches the finish line, it turns around and flies directly back to the runner. What is the cumulative distance Lb the bird travels?
Time to finish t = Lr/Vr = 7.1/2.2 h
Lb = Vb*t = 4.4*7.1/2.2 = 14.2 km
- Andrew SmithLv 711 months ago
This is a standard trick question.
The runner finishes in a time of 7.1/ 2.2 hr
and in that time the bird will have flown v* t = 4.4 * 7.1/2.2 = 14.2 km
Big bird pointed out that as written the bird only does ONE return trip.
To do that consider that the bird took a time t for the trip.
In that time the runner moves forwards 2.2t
so that the bird must have flown 7.1 +(7.1-2.2t) km = 14.2-2.2t
The time taken must be distance / speed
so t = (14.2 - 2.2t) / 4.4
4.4t = 14.2-2.2t
6.6 t = 14.2
t = 14.2/6.6
the distance the BIRD flew must be 4.4 t = 4.4* 14.2/6.6 = 2/3 * 14.2 = 9.5km
Now you COULD Use this as the basis of an infinite sum of a geometric progression if you can remember your fifth form maths. To find the distance flown after each iteration.
Why not do that as a simple practice exercise?