Two jets leave Ontario at the same time, one flying east at aspeed 20km/h greater than the other. Which is flying west. After 4 hours, the planes are 6000 km apart. Find their speeds

Question

I just need this solved please show steps too. Thanks in advance.

Answer 1

plane 1 is flying east at speed (x+20)

plane 2 is flying west at speed x

distance = speed * time

4 hours later:

plane 1 is 4(x+20) km east of the starting point.

plane 2 is 4x km west of the starting point.

add these to get the total distance

4(x+20)+4x

simplify:

4x+80+4x

8x+80

now set this equal to the given distance: 6000 km

8x+80=6000

subtract 80 from both sides

8x=5920

divide by 8

x=740

Answer:

Plane 1 is flying east at 760 km/h

Plane 2 is flying west at 740 km/h

Answer 2

j1=e; vel+20kph

t2=w; vel; kph

t=4hrs

side "c"=6000km

I'm using a trig formula: a^2+b^2=c^2.

t^2+(t+20)^2=36*10^6

t2=537.635/4hrs=134.4087.........km/hr

t1=557.635............km/hr

Answer 3

1. Sketch a quick pic: <-----(x)-------•-------(x+20) ---->
2. 4 hours later the distance is 6000, so you would multiply each of the distances by 4 to get 6000; so 6000 = 4(x) + 4(x+20); do distributive property and solve for x
3. 6000 = 4x + 4x +80; 8x = 5920, so x = 740 km/hr

Answer 4

Let re= speed of east bound plane = rw+20. rw= speed of west bound plane

d=rt, distance = rate of speed times time

6000= 4rw 4(rw+20)

8rw = 6000-80

rw = 5920/8 = 740 km/hr

r3 = 740+20 = 760 km/hr

check the answer, substituting into d=rt, d=4re + 4rw

6000= 4(760) + 4(740) = 4(760+740) = 4(1500) = 6000