Question

Plane A leaves Tulsa at 2:00 p.m., averaging 300 mph and flying in a northerly direction. Plane B leaves Tulsa at 2:30 p.m., averaging 225 mph and flying due east. At 5:00 p.m., how far apart will the planes be?

Answer 1

Answer 1061

Step-by-step explanation:

Answer 2

Answer:

At 5:00 p.m. the planes will be $1061.323$ miles apart.

Step-by-step explanation:

To solve this problem, I add a picture of the situation.

We know that speed is distance over time  ⇒

$Speed=\frac{distance}{time}$ (I)

The first step to solve this exercise is to graph the situation. We can draw a right triangle which vertices will be ''Tulsa'', and the planes ''A'' and ''B'' at 5:00 p.m.

In order to know the measures of the sides, we are going to calculate them using the equation (I)

Plane A leaves Tulsa at 2:00 p.m.

Therefore, at 5:00 p.m. it will have flown 3 hours ⇒

$300\frac{mi}{h}=\frac{distance}{3h}$ ⇒

$distance=(300\frac{mi}{h}).(3h)$

$distance=900mi$

At 5:00 p.m. the distance from the plane A to Tulsa is 900 mi

Plane B leaves Tulsa at 2:30 p.m.

Therefore, at 5:00 p.m. it will have flown 2.5 hours ⇒

$225\frac{mi}{h}=\frac{distance}{2.5h}$

$distance=(225\frac{mi}{h}).(2.5h)$

$distance=562.5mi$

At 5:00 p.m. the distance from the plane B to Tulsa is 562.5 mi

Finally, we can find the distance between the plane A and the plane B using the Pythagorean theorem :

$(900mi)^{2}+(562.5mi)^{2}=(Distance_{A-B})^{2}$

$810000mi^{2}+316406.25mi^{2}=(Distance_{A-B})^{2}$

$(Distance_{A-B})^{2}=1126406.25mi^{2}$

$Distance_{A-B}=\sqrt{1126406.25mi^{2}}$

$Distance_{A-B}=1061.322877mi$ ≅ $1061.323mi$

At 5:00 p.m. the planes will be 1061.323 miles apart.