Question 1165283: The velocity of an airplane in still air is 130kph. The velocity of the airplane due east is 25 kph. If the plane travels east and returns back to its base again after 4 hours. At what distance does the plane travel due east?
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
You have misstated the problem. The speed of the airplane, i.e. the magnitude of the velocity of the airplane in still air is 130 km/h.
If the aircraft managed 25 km/h going east, then there must have been a wind from the east at 130 minus 25 = 105 km/h. Consequently, on the return trip, the aircraft must have had a resultant velocity of 105 + 130 = 235 km/h.
Unless someone picked up the airplane's base and moved it while the plane was in the air, the distance on the outbound trip has to be equal to the distance of the return trip. Since distance is equal to rate times time and the elapsed time of the return trip must be the remainder of the four hour total that was not used on the outbound trip, the following must hold:
)
Then calculate 
Or, if you don't feel like doing all of that arithmetic, you could confidently assert that the distance travelled due east is the same as the distance travelled due west. I'm not sure how well that would be received by your professor, teacher, or instructor, but it is a demonstrably true assertion that exactly answers the question asked.
John
My calculator said it, I believe it, that settles it
I > Ø
|
|
|
