An airplane flies from its headquarters at night to a city 510 miles away and returns back the next morning. The total flying time for the round-trip flight is 3.9 h. The plane travels the first half of the trip at 255 mph with no wind. How strong is the wind on the return flight?

13 miles per hour

Step-by-step explanation:

If the first half of the trip was travelled at the speed of 255 mph and the distance from the headquarters to the city is 510 miles, the time of travel for the first half is:

255 miles 1 hour

510 miles ? hours

= 510/255 × 1/1

= 2 hours

So, since the first half of the journey was at the speed of 255 miles per hour, therefore it took the plane 2 hours to get to the city from the headquarters (covering a distance of 510 miles).

But, if the time for the entire round trip is 3.9 hours, the return time is therefore:

3.9 hours 2 hours

= 1.9 hours( so this is the return time)

Since returning will still entail covering the same distance of 510 miles, then the speed of return is:

Speed = distance/time

= 510/1.9 = 268.42 mph.

To determine how strong the wind was on the return, we subtract the speed of the first half of the trip from the speed of the return flight:

= 268.42 - 255

= 13 miles per hour (13 mph)

i don't know, but that looks pretty bad. my school passed us unless we failed in all the subjects, but idk abt ur school's rules.

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