Question

An Air Force plane left Singapore and flew toward the maintenance facility at an average speed of 150 mph. A cargo plane left some time later flying in the same direction at an average speed of 180 mph. After flying for five hours the cargo plane caught up with the Air Force plane. How long did the Air Force plane fly before the cargo plane caught up?

Answer 1

I got the answer of 30mph, but the answer key says 6mph. Can someone guide me on how it is 6mph?

Answer 2

@Nnesha

Answer 3

@jim_thompson5910

Answer 4

the distance of the cargo at some time (d = rate*time), will have less traveling time than the air force call it a delta t

air force d=rate*t
cargo d = rate *(t- delta t)

when elapsed time t is 5 hours, they are the same distance, find the delta t, the stall period

Answer 5

i used a rate, time, distance table

Answer 6

i got it now actually 180x5=900
900/150=6

Answer 7

thank you though!!

Answer 8

yeah you can do calculations like the air force traveled 5 hours at that rate, total distance is 750....

or since here we are looking for when they are at the same place at the same time, it is not to bad

d = r*t that is it

the distance , given as the same for both, at t=5 hours, is just a bit different for the cargo

t = total elapsed time for both
T = cargo wait to fly time

d=r*t for the airforce
but
d = r*(t - T) for cargo, he was traveling at that rate for that T time less, still covered same distance in 5 hours , so just figure the T

Answer 9

i see , yeah the table lays it out, but you put a time variable in for both which is ok

air force - distance = 150*t

but, why di dyou put t - 5 for the cargo travel time,

Answer 10

For the rate*time, think of expanding it out, D=r*(t - T) = rate*total time - rate*wait time

if you wait to fly, your distance you would have had, rate*time, is reduced by that rate *the wait time