Question

A concorde files 2000km at a speed of 1600km/h and returns at a speed of 1000km/h. Find the average speed for the who trip???

Answer 1

average speed is not the average of the speeds. calculate total distance (4000) and divide by total time. you need time going and time returning

Answer 2

time going is
\[\frac{2000}{1600}=1.25\] time returning is
\[\frac{2000}{1000}=2\]
total time is
\[1.25+2=3.25\] so average speed is
\[\frac{4000}{3.25}\]whatever that is

Answer 3

I followed the very same steps, all I did different was change the 1.25 in 1 hour and 15 mins

Answer 4

60* .25

Answer 5

well i would keep the units in hours until the end

Answer 6

You will want to keep the time units in hours (not hours and minutes) because you are looking for km/hour
after finding 1.25 hours one way, you need the time to come back. Total distance divided by total time is the average speed. See @sat above

Answer 7

you do not want to divide by 3 hours and fifteen minutes. you want to divide by 3.25

Answer 8

bless you guys

Answer 9

i get 1230.77 mph rounded
btw you might notice that this average is
\[\frac{4000}{\frac{2000}{1600}+\frac{2000}{1000}}\] replacing 4000 by 2m you get
\[\frac{2m}{\frac{m}{1600}+\frac{m}{1000}}=\frac{2}{\frac{1}{1600}+\frac{1}{1000}}=1230.77\]
meaning that the distance traveled in this problem was in fact irrelevant

Answer 10

and not to bore you to death with this problem, but if you have two numbers
\[x_1,x_2\] the "harmonic mean" defined as
\[\frac{2}{\frac{1}{x_1}+\frac{1}{x_2}}\]