Two planes start from LAIA and fly in opposite directions. The 2nd plane starts 1/2 hour after the 1st plane, but its speed in 80 kilometers per hour faster. Find the airspeed of each plane if 2 hours after the first plane departs the planes are 3200 kilometers apart.
Explanation needed please
2007-06-16 04:39:51 · 5 answers · asked by Lollipop 3 in Science & Mathematics ➔ Mathematics
Speed of plane 1 = x : speed of plane 2 = x+80
Plane 1 travels 2x in those 2 hours....
Plane 2 travels 1.5(x+80) in those 2 hrs.....
Therefor: 1.5(x+80) + 2x = 3200
Simplify..... 1.5x + 120 + 2x =3200......
3.5x=3080
x=880.....
Speed of plane 1 = 880 MPH
Speed of plane 2 = 960 MPH
Got it?
2007-06-16 04:59:40 · answer #1 · answered by S. E. Charles 3 · 1⤊ 0⤋
Let the speed of the first plane be x kmph.Hence the speed of the second plane is x+80 kmph
The first plane starts half an hour earlier.Hence it has alrady travelled a distance of x/2 km.
as the two planes are moving in opposite directions,their relative is sped is x+x+80 or 2x+80 km
In 3/2 hrs the distance between them would be 3/2(2x+80) or 3x+120 km plus x/2 km travelled by the first plane due to earlier take-off
by the problem,
3x+120+x/2=3200
=>6x+240+x=6400[multiplying both sides by 2]
=>7x=6400-240=6160
=>x=6160/7=880
therefore the speed of the first plane is 880 kmph and that of the second one is 880+80=960 kmph
2007-06-16 04:53:47 · answer #2 · answered by alpha 7 · 0⤊ 2⤋
Let, speed of the 1st plane be : x km/hr
and speed of the 2nd plane is : x+80 km/hr
Distance of 1st plane in 2hrs is : 2x km
and that of 2nd plane in 2hrs is : 3/2(x+80) km
Hence, 2x+3/2(x+80) = 3200
=> 2x+3/2x+120 = 3200
=> 7/2x = 3080
=> x = 6160/7
Therefore, x = 880
and, x+80 = 960
Therefore, speed of the 1st plane would be 880 km/hr and speed of the 2nd plane would be 960 km/hr.
Is that clear enough??!!!
2007-06-23 19:18:29 · answer #3 · answered by gothic_vampy_666 1 · 0⤊ 0⤋
Let x = speed of the 1st, x + 80 speed of the 2nd
2 = hours traveled by 1st, 1.5 = hours traveled by 2nd
Equation:
2x + (1.5 * [x + 80]) = 3200
2x + 1.5x + 120 = 3200
3.5x = 3200 - 120
3.5x = 3080
x = 3080 / 3.5
x = 880
Speed: 1st = 880 km/hr, 2nd = 880 + 80 or 960 km/hr
Proof of distance traveled:
1st = 880 * 2 or 1760 km; 2nd = 960 * 1.5 or 1440 km
Total distance traveled 1760 + 1440 or 3200 km
2007-06-21 19:57:11 · answer #4 · answered by Jun Agruda 7 · 2⤊ 0⤋
let T= time of 2nd plane
V=velocity of first plane
distance= 3200
plane 1 distance: [(T + .5) x V]
plane 2 distance: [T x (V+80)]
therefore: 3200 = TV + .5V + TV + 80T, or
3200 = 2TV + .5V + 80T
since Time = "1.5" hours (2 - .5), it becomes
3200 = 3V + .5V +120, or
3200 = 3.5V + 120, or
3080 = 3.5 V, or
V = 880 km/hr (first plane)
and
V = 880 + 80, or 960 km/hr (second plane)
Checking the answer:
(880 x 2) + (960 x 1.5) = 1760 + 1440 = 3200
2007-06-23 12:53:50 · answer #5 · answered by Kevin S 7 · 0⤊ 0⤋