210k views
2 votes
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

User Giordano
by
8.4k points

2 Answers

1 vote
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.
User Hidar
by
8.1k points
2 votes
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
User Yee Liu
by
8.2k points