An aeroplane has to go along straight line from A to B, and back again. The relative speed with respect
to wind is V. The wind blows perpendicular to line AB with speed v. The distance between A and B is l.
The total time for the round trip is :
Answers
Answer:
The time (distance divided by speed) in the downwind half is
1/ V +v
And the time in the upwind half is
1/ V-v
So, the total time is (1/V+v) + (1/V-v)
Working toward a common denominator,
{(V-v) + (V+v)} / (V-v) (V+v)
Then,
2 V / V^2 + V v - V v -v^2
Combining like terms,
2 V / V^2 -v^2
If V = 1, and v = .1, when AB = 1,
2 / 1 - .01
2 / .99
Total time is 2.02 units
To put this into a real-world scenario:
Instead of AB being 1,
AB= 100 miles
V=100 mph
v=10 mph
Downwind time: 100/110
Upwind time: 100/90
Total time: (100/110) + (100/90)
(9000/9900) + (11000/9900)
20,000/9900
Total time is 2.02 hours
This is not the 2.00 hours found by doubling the distance and assuming the winds cancel out.
Answer:
From the relative velocity concept:
when going from A to B time taken=>
t1 = l/(V-v)
from B to A while returning,
t2 = l /(V+v)
Total Time = t1 + t2 => l /(V+v) + l /(V-v) = > l/(1/(V-v) + 1/(V+v)) seconds.
Hope it's help you.....