Question

A ship traveled for 4 hours heading west and for 5 hours heading north. If the total distance traveled was 119 miles, and the ship traveled 5 miles per hour faster west, at what speed was the ship traveling west

Answer 1

To answer this problem, we will use the formula of hypotenuse:

c^2 = a^2 + b^2

Where,

c = total distance traveled / displacement = 119 miles

a = distance traveled north = 5 hours * x mph =  5 x

b = distance traveled west = 4 hours * (x + 5) mph = 4 (x + 5)

Substituting to the equation:

119^2 = (5 x)^2 + [4(x + 5)]^2

14,161 = 25 x^2 + 16 (x + 5)^2

14,161 = 25 x^2 + 16 (x^2 + 10 x + 25)

14,161 = 25 x^2 + 16 x^2 + 160 x + 400

0 = 41 x^2 + 160 x – 13,761

x =[ - b ± sqrt (b^2 – 4ac)]/ 2a

x = [- 160 ± sqrt (160^2 – 4 * 41 * (– 13,761))] / 2 * 41

x = - 1.95 ± 18.42

x = -20.37 , 16.47

Since speed cannot be negative, therefore x = 16.47

Therefore the speed of the ship travelling west is:

x + 5 = 21.47 mph         (ANSWER)