Two ships leave Whittier, AK at 7 AM.
- The first ship sails towards Valdez on a 54° course at a constant speed of 36 mi/h.
- The second ship sails towards Knight Island on a 144° course at a constant speed of 42 mi/h.
Find the distance between the ships at 11 AM.
Answer (3)
11 A M − 7 A M = 4 h a = 36 mi / h ⋅ 4 h = 36 ⋅ 4 mi l es = 144 mi l es b = 42 mi / h ⋅ 4 h = 42 ⋅ 4 mi l es = 168 mi l es α = 14 4 o − 5 4 o = 9 0 o x 2 = 14 4 2 + 16 8 2 − 2 ⋅ 144 ⋅ 168 ⋅ cos 9 0 o x 2 = 20736 + 28224 − 2 ⋅ 144 ⋅ 168 ⋅ 0 x 2 = 48960 x = 48960 x ≈ 221.27 ( mi l es )
or P y t ha g or a s t h eore m : x 2 = 14 4 2 + 16 8 2 x ≈ 221.27 ( mi l es ) A n s w er : 221.27 mi l es .
t = 11 − 7 = 4 [ h ] ∧ S = V ⋅ t S 1 = V 1 ⋅ 4 = 36 ⋅ 4 = 144 [ mi ] an d S 2 = V 2 ⋅ 4 = 42 ⋅ 4 = 168 [ mi ] x 2 = ( S 1 ) 2 + ( S 2 ) 2 ⇒ x 2 = 14 4 2 + 16 8 2 ⇒ x 2 = 48960 x ≈ 221.3 [ mi ] A n s . T h e d i s t an ce b e tw ee n t h e s hi p s a t 11 A M i s 221.3 mi
The distance between the two ships at 11 AM is approximately 221.27 miles. This is calculated using the Law of Cosines based on their distances traveled and the angle between their courses. After 4 hours, the first ship travels 144 miles and the second travels 168 miles.
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