A passenger plane and a jet leave att the same time from the same airport,traveling in opposite directions. The jet travels at twice the speed of the passenger plane. they are 2400 mi apart in 4 hours. Find the speed of the passenger plane and jet.

speed of passenger ---- x mph

speed of jet -----------2x mph

they each went for 4 hours, so
distance covered by jet = 4(2x) = 8x
distance covered by pass = 4x

8x + 4x = 2400
12x = 2400
x = 200

passenger plane went 200 mph, and the jet went 400 mph

Just noticed that I answered the same question back in Nov of 2007
http://www.jiskha.com/display.cgi?id=1194411287

To find the speed of the passenger plane and the jet, we can set up a system of equations based on the information given.

Let's denote the speed of the passenger plane as 'x' mph, and the speed of the jet as '2x' mph since it travels at twice the speed of the passenger plane.

Since both planes are traveling in opposite directions, we can add their speeds to find their relative speed.

So, the relative speed of the two planes is 'x + 2x = 3x' mph.

Now, we know that they are 2400 miles apart and they traveled for 4 hours. Therefore, using the formula: Distance = Speed × Time, we can set up the equation:

2400 = 3x × 4

Simplifying this equation, we get:

2400 = 12x

Dividing both sides of the equation by 12, we find:

x = 200

So, the speed of the passenger plane is 200 mph.

Since the speed of the jet is twice the speed of the passenger plane, the speed of the jet is 2 * 200 = 400 mph.

Therefore, the speed of the passenger plane is 200 mph, and the speed of the jet is 400 mph.