To solve the problem, first we need the coordinates of the points involved:
- Starting Point: \( S(-6, 8) \)
- Obstacle 1: \( O1(-6, 0) \)
- Obstacle 2: \( O2(0, 0) \)
Part A: Find the distance from the starting point to obstacle 2.
We can use the distance formula to find the distance between two points in a Cartesian coordinate system, which is given by: \[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
For the points \( S(-6, 8) \) (Starting Point) and \( O2(0, 0) \) (Obstacle 2):
- \( (x_1, y_1) = (-6, 8) \)
- \( (x_2, y_2) = (0, 0) \)
Plugging these coordinates into the distance formula: \[ d = \sqrt{(0 - (-6))^2 + (0 - 8)^2} \] \[ d = \sqrt{(0 + 6)^2 + (-8)^2} \] \[ d = \sqrt{(6)^2 + (8)^2} \] \[ d = \sqrt{36 + 64} \] \[ d = \sqrt{100} \] \[ d = 10 \]
So, the distance from the starting point to obstacle 2 is 10 meters.
Part B: How many meters is one full lap around the course?
The course forms a right triangle, and we will calculate the lengths of each side.
-
Length of Side 1 (from Starting Point S to Obstacle 1 O1):
- Using the coordinates \( S(-6, 8) \) and \( O1(-6, 0) \): \[ d = \sqrt{(-6 - (-6))^2 + (8 - 0)^2} \] \[ d = \sqrt{(0)^2 + (8)^2} \] \[ d = \sqrt{64} \] \[ d = 8 \text{ meters} \]
-
Length of Side 2 (from Obstacle 1 O1 to Obstacle 2 O2):
- Using coordinates \( O1(-6, 0) \) and \( O2(0, 0) \): \[ d = \sqrt{(0 - (-6))^2 + (0 - 0)^2} \] \[ d = \sqrt{(6)^2 + (0)^2} \] \[ d = \sqrt{36} \] \[ d = 6 \text{ meters} \]
-
Length of Side 3 (from Obstacle 2 O2 back to Starting Point S):
- From Part A, we previously calculated this distance as \( 10 \) meters.
Now we add up all the sides to get the total distance for one full lap around the course: \[ \text{Total distance} = \text{Side 1} + \text{Side 2} + \text{Side 3} \] \[ \text{Total distance} = 8 + 6 + 10 \] \[ \text{Total distance} = 24 \text{ meters} \]
So, one full lap around the course is 24 meters.