At 1000 hours an aircraft starts from A to fly to B, a flight which will take 4 hours. At 1042 a second aircraft starts the same journey and will complete the flight in 2 hours 30 minutes. If the distance from A to B is 800 miles, at what time will the aircraft be the same distance from B?
when the distance between the 2 planes = 0 then they are the same distance from pointB
800/4 is the speed of the 1st aircraft 800/(2.5) is the speed of the second aircraft..
200 mph and 320 mph sound about right...
does 1042 means 42 minutes after 10?
yeah I got the speeds
thats right 45 mins after 10
ok, so if we convert 45 minutes into a fraction of an hour, we will be able to work with it better.... 3/4 or .75 of an hour....
42 mins* but thats where I got stuck
ok... 60/42 is a good enough fraction to work with :)
yeah got that
divide each by 3... 20/14....10/7.... 15/7 seems good... you agree?
10/7 seems good
yeah 10/7 looks ok
these are the options given: a 1110 hrs b 1124 hrs c 1142 hrs d 1152 hrs
lets make some equations then... 200x + 10/7 = D 320x = D does that make sense to you?
I think i missed it... we need to know hoe far the first plane has traveled and add that...not the hours...lol
200(10/7) is the "added" distance, or the advantage that the 1st plane has...
ermm .. Y would it be + 10/7. Dont get how the + comes into it
ok that makes sense
200x + 285.7 = D 320x = D where D is the equal distance flown .... hows that?
yeah that makes sense
makes better sense, but what doesnt make sense to me yet is that it is less than an hour, but the 1st plane has gone more than 200? gotta see where I missed that at :)
hmm ...
200(60/42) = 190.5 looks alot better, musta got it turned around in me head somewhere :)
I didnt get 190.5
7/10 should be our multiplier... 60 is the whole, 42 is the part; part/whole = percent.... .7
200(.7)= 140... that HAS to be it lol maybe
One way to consider it: Look at the time when only the first is flying; it is going to 200mph for 42 minutes, so will end up 140miles away from A. NEXT, look at the motion after they are both moving. If you apply a vector of -200mph to both planes, the one that set off first is effectively stationary, so you only need to consider how long it takes the second plane to 'catch up' the 140 mile head start at (320 - 200)mph = 120mph, which is just time = distance over speed. Add the initial 42 minutes and you're done.
200/60 tells us miles per minute; times 42 equals distance traved in 42 minutes...
200x+140 = 320x 140 = 320x - 200x 140 = 120x x = 140/120 = 14/12 = 7/6... 1 hour and 6/60 minutes.... if it agrees with Newtons :)
oh that make sense so it should be 200(42/60) = 140 (I.e the distances travelled by plane A)
Yep.... 1042 + 0106 = 1148...
but these are the options: a 1110 hrs b 1124 hrs c 1142 hrs d 1152 hrs
1 hour and 1/6 of another .... ack!! my math is off today...
1hour and 10 minutes... double check that :)
1042+0110 = 1152....
But yes, I agree. Time = 42 minutes + (140/120) minutes = 42 + (7/6) hours = the same as yours.
Ahh .. thanks you
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