an orienteer runs 300m north and then 400m to the southeast (at a 45 degree angle to north). if he runs at a constant speed, how far away is he from the starting position?
hvk
To find the distance from the starting position, we can use the Pythagorean theorem.
The orienteer runs 300 meters north, which means he has travelled 300 meters vertically (in the y-direction). Then he runs 400 meters to the southeast at a 45-degree angle, which can be split into two equal distances of 400/sqrt(2) meters in both the x-direction and y-direction.
Let's calculate the distance in the x-direction and y-direction:
Distance in the x-direction = 400/sqrt(2) * cos(45 degrees) = 400/sqrt(2) * 1/sqrt(2) = 400/2 = 200 meters
Distance in the y-direction = 400/sqrt(2) * sin(45 degrees) = 400/sqrt(2) * 1/sqrt(2) = 400/2 = 200 meters
Now we can use the Pythagorean theorem to find the distance from the starting position:
Distance = sqrt((300 + 200)^2 + (200)^2) = sqrt(500^2 + 200^2)
= sqrt(250000 + 40000)
= sqrt(290000)
≈ 539.75 meters
Therefore, the orienteer is approximately 539.75 meters away from the starting position.