Question

1. airplane speed two planes start from los angeles international airport and fly in opposite directions. the second plane starts \\(\frac{1}{2}\\) hour after the first plane, but its speed is 80 kilometers per hour faster. two hours after the first plane departs, the planes are 3200 kilometers apart. find the airspeed of each plane.

Explanation:

Step1: Define variables

Let speed of first plane be $x$ km/h.

Step2: Express distances

$d_1 = 2.5x$, $d_2=2(x + 80)$

Step3: Set up equation

$2.5x+2(x + 80)=3200$

Step4: Simplify equation

$4.5x+160 = 3200$

Step5: Isolate variable term

$4.5x=3040$

Step6: Solve for $x$

$x=\frac{3040}{4.5}$

Step7: Find speed of second plane

$x + 80$

Answer:

Let the speed of the first - plane be $x$ km/h. Then the speed of the second - plane is $(x + 80)$ km/h.

The first plane flies for $t_1=2.5$ hours and the second plane flies for $t_2 = 2$ hours.

The distance formula is $d=v\times t$ (where $d$ is distance, $v$ is speed, and $t$ is time).

The distance covered by the first plane, $d_1=x\times2.5 = 2.5x$ km.

The distance covered by the second plane, $d_2=(x + 80)\times2=2x+160$ km.

Since the sum of their distances is 3200 km, we have the equation:

$2.5x+2x + 160=3200$

$4.5x+160 = 3200$

$4.5x=3200 - 160$

$4.5x=3040$

$x=\frac{3040}{4.5}=\frac{30400}{45}=\frac{6080}{9}\approx675.56$ km/h

The speed of the first plane is $x=\frac{6080}{9}\approx675.56$ km/h

The speed of the second plane is $x + 80=\frac{6080}{9}+80=\frac{6080+720}{9}=\frac{6800}{9}\approx755.56$ km/h