Plane A leaves Albany at 2:00 p. m., averaging 400 mph and flying in a northerly direction. Plane B leaves Albany at 2:30 p. m., averaging 325 mph and flying due east. At 5:00 p. m., how far apart will the planes be? A. 1,449 B. 1,649 C. 1,849

From 2:00 pm to 5:00 pm, Plane A traveled for 3 hours. The total distance it had traveled is the product of its speed and the number of hours which is,

400 mph x 3 hours = 1200 miles

For Plane B, from 2:30 pm to 5:00 pm, it traveled for 2.5 hours and the total distance it had traveled is 812.5 miles. The unknown distance is the hypotenuse of the right triangle such that,

x² = (1200 miles) ² + (812.5 miles) ²

The value of x from the equation is 1449.19 miles. Thus, the answer is letter A.

Plane A traveled for 3 hours and total distance it had traveled is:

400 mph x 3 hours = 1200 miles (using D = V x t)

Plane B traveled for 2.5 hours and the total distance traveled is 812.5 miles

Using Pythagoras theorem;

x² = (1200) ² + (812.5) ²

x = 1449.19 miles

Hence, the best answer is option A. 1,449