A jet leaves an airport traveling at a steady rate of 600 km/h. Another jet leaves the same airport 3/4 hours later traveling at 800km/h in the same direction. How long will it take the second jet to overtake the first?
DO NOT SOLVE IT! JUST SHOW ME HOW TO SET IT UP AND WHY YOU SET IT UP LIKE THAT!
2007-02-04 02:37:51 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
by the time the second jet leaves, the first jet has already traveled 3/4 * 600 km. (time * velocity = distance). let us say that it takes the second jet T hours to catch the first jet. in that time, the the second jet will travel T * 800 km and the first jet will travel T * 600 km. remember that the first jet had a head start of 3/4 hours during which it traveled 3/4 * 600 km. when the second jet catches the first jet, they will have both traveled the same distance (that's what it means to catch up). so...
total distance first jet travels = (3/4 * 600) + (600 * T)
total distance second jet travels = 800 * T
since they both travel the same distance,
800 * T = (3/4 * 600) + (600 * T)
now you can solve for T
another way to solve this problem is to realize that the second jet is catching up to the first jet at a rate of 200 km/h. the first jet still had a head start of 3/4 hours and traveled 3/4 * 600 km. the time T it takes can be found by dividing the head start distance by this relative velocity:
200 * T = 3/4 * 600
T = (3/4 * 600) / 200
2007-02-04 03:04:47 · answer #1 · answered by michaell 6 · 0⤊ 0⤋
Jet 1 has speed = 600 km / h
Jet 2 has speed = 800 km / h
Relative speed of jet 2 to jet 1 is 200 km / h
ie can assume Jet 1 stopped and Jet 2 travelling at 200 km / h
In 45 mins Jet 1 will have travelled a distance of 3/4 x 600 km = 450 km
Question then becomes "How long will it take for Jet 2 to travel 450 km at 200 km / h?"
As you requested , I have not calculated an answer.
2007-02-04 03:13:33 · answer #2 · answered by Como 7 · 0⤊ 0⤋
You could use distance = rate times time, and since the distance is trhe same (they left from the same place and are together when one catches up) you have rt from one plane = rt from the other
The slower plane has 3/4 extra hours so its time is t + 3/4; the faster plane is t because that's what they asked for.
Another way would be to see how much of a head start the first plane had by 3/4 times 600, then see how long it would take the faster plane to catch up, since it gains 200 km/h.
2007-02-04 02:46:16 · answer #3 · answered by hayharbr 7 · 0⤊ 0⤋
Distance is speed * time.
So you set their distances equal, because that's when they're next to each other, and the faster place will pass the other.
Distance = 600 * t = 800 * (t-3/4)
600t = 800t - 600
-200t = -600
And so on... I just remembered NOT to solve...
You use t-(3/4) because it flies the same hours as the other (t) minus 3/4 hours because it flew after.
2007-02-04 02:53:05 · answer #4 · answered by Anonymous · 0⤊ 0⤋
OK, you take 600 x 0.75. . . That will equal the distance the first plane is ahead of the second one.
The second one is traveling at 800 - 600. . . Then you know how much faster the second plane is traveling.
Divide your first answer by your second answer, and you will have the time. . .
2007-02-04 02:45:45 · answer #5 · answered by Walking Man 6 · 0⤊ 0⤋
time taken=distance to be closed by the relativespeed
distance to be closed=dist travelled by jet1 in 3/4 hr@600kph
relative speed is the differencein speeds as they are travelling in thesame direction
=[(3/4)*600]/200
2007-02-04 02:43:52 · answer #6 · answered by raj 7 · 0⤊ 0⤋
The variable that is common to both jets is distance
(let`s call it X)
The formula for distance at constant speed is:
X = V * t
Since X is the same for both jets (t is the time for the second jet):
Va * (t + 3/4h) = Vb * t
From here you only have to resolve for t.
2007-02-04 03:12:19 · answer #7 · answered by gquiroz2005 2 · 0⤊ 0⤋