an rcmp patrol boat left long beach, british columbia, and sailed west for 2h at 15km/h, then north for 2h at 20km/h, then east for 1h at 10km/h. How far was the patrol boat from long beach, to the nearest tenth of a kilometre?
2007-06-02 10:29:37 · 2 answers · asked by harleen 1 in Education & Reference ➔ Homework Help
The boat is 44.7 km from Long Beach.
Since the boat traveled 2 hours at 15 km/h initially, it was 2 h x 15 km/h = 30 km west of Long Beach when it turned north. It traveled north a distance of 40 km (2 h x 20 km/h = 40 km). Then it traveled back east a distance of 10 km (1 h x 10 km/h = 10 km).
To find the total distance it is west of Long Beach when the boat stopped traveling, we can subtract 10 km from 30 km, because an easterly direction is directly opposite a westerly direction. So the boat is 20 km west of Long Beach at the end of this journey. The boat's northerly distance from Long Beach hasn't changed. That means we can use the Pythagorean Theorem to find the diagonal distance of the boat from Long Beach.
D = √(40 km)² + (20 km)²
D = √(1600 km² + 400 km²)
D = √2000 km²
D = √[(400 x 5) km²]
D = 20 √5 km
D ~ 20 x 2.236 km
D ~ 44.72 km
D ~ 44.7 km to the nearest tenth.
2007-06-02 10:36:20 · answer #1 · answered by MathBioMajor 7 · 1⤊ 0⤋
The boat sailed 2*15 = 30km west, then 2*20 = 40km north, and then 10km east.
If the boat is x km. from Long Beach, then:
x^2 = 40^2 + 20^2
= 1600 + 400
= 2000
= 400 * 5
x = sqrt(400)sqrt(5)
= 20sqrt(5)
= 44.7km. to nearest 0.1 km.
2007-06-02 17:45:26 · answer #2 · answered by Anonymous · 1⤊ 0⤋