Sure, let's tackle each part of the problem step by step.
### a)
The distance that the runner covers if the start and the finish line are the same is the circumference of the circular track. The formula for the circumference of a circle is given by:
[tex]\[ \text{Circumference} = 2 \pi r \][/tex]
where [tex]\( \pi \approx 3.14159 \)[/tex] and [tex]\( r \)[/tex] is the radius of the circle.
Given:
- Radius [tex]\( r = 80 \)[/tex] meters
Thus,
[tex]\[ \text{Circumference} = 2 \times \pi \times 80 \][/tex]
[tex]\[ \text{Circumference} = 2 \times 3.14159 \times 80 \][/tex]
[tex]\[ \text{Circumference} \approx 502.65 \text{ meters} \][/tex]
So, the distance covered by the runner is approximately 502.65 meters.
### b)
The displacement is the straight-line distance from the starting point to the ending point. Since the runner returns to the starting point after one complete circuit, the displacement is:
[tex]\[ \text{Displacement} = 0 \text{ meters} \][/tex]
### c)
The average speed is calculated as the total distance covered divided by the total time taken. The formula for average speed is:
[tex]\[ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \][/tex]
Given:
- Total distance (Circumference) [tex]\( = 502.65 \)[/tex] meters
- Total time [tex]\( = 60 \)[/tex] seconds
Thus,
[tex]\[ \text{Average Speed} = \frac{502.65}{60} \][/tex]
[tex]\[ \text{Average Speed} \approx 8.38 \text{ meters/second} \][/tex]
### d)
The average velocity is calculated as the total displacement divided by the total time taken. For half the circuit, the formula for average velocity is:
[tex]\[ \text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}} \][/tex]
If the runner races halfway around the track, they cover half of the circumference and the displacement would be the straight-line distance between the starting point and the point directly opposite on the circle.
However, in this context, we look at the displacement along the track making it easier:
- Half of the total distance (Half-Circumference) [tex]\( = \frac{502.65}{2} \approx 251.33 \)[/tex] meters
- Total time for half the circuit [tex]\( = 30 \)[/tex] seconds
Thus,
[tex]\[ \text{Average Velocity} = \frac{251.33}{30} \][/tex]
[tex]\[ \text{Average Velocity} \approx 8.38 \text{ meters/second} \][/tex]
So the average velocity for half the track is approximately 8.38 meters/second.
