ok so i got the answer but can someone help me explain it please!
On a flight from Sydney, Nova Scotia to Saskatoon, Sask. the plane averaged 760km/hr. On the return trip from Saskatoon to Sydney, the same plane averaged 904km/hr. The distance from Sydney to Saskatoon is approximately 4112km. How much longer was the trip West than the trip East? How long was each flight in minutes?
ANSWER=Time = 4112/760 = 5.4 hrs ; 5.4hrs * 60 = 324 mins
Time = 4112/904 = 4.55 hrs = 273 mins
Sydney to Saskatoon = 324 mins
Saskatoon to Sydney = 273 mins
The trip West is 51 minutes longer
2007-09-06 14:01:05 · 3 answers · asked by sweethoney9079 4 in Science & Mathematics ➔ Mathematics
On the outward journey the plane's average speed was760 km/h which is 760/60 ot 38/3 km/minute
therefore time in minutes taken for the outward journey
=4112 divided by 38/3
=4112*3/38
=324.63 minutes
On the return journey,the average speed of the plane was 904 km/hr which is equal to
904/60 or 226/15 km/minute
Therefore time taken for the return journey
=4112 divided by 226/15
=4112*15/226
=272.92 minutes
Therefore out ward journey took 324.63-272.92 or 51.71 minutes longer
2007-09-06 14:19:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
i divided the amount of time from the trip from Sydney to Saskatoon then i divided the arrival time by the trip that is how i got my answer for the minute time also i subtracted the minutes from Sydney to Saskatoon from the minutes from Saskatoon to Sydney that is how i got 51 minutes longer
2007-09-06 21:11:55 · answer #2 · answered by Anonymous · 0⤊ 0⤋
What's to explain;-} you have head-winds going west.
Then you get tail-winds flying east.
Assuming the winds were constant, how fast were they?
2007-09-06 21:08:44 · answer #3 · answered by Robert S 7 · 0⤊ 0⤋