An athlete, in a race, accelerates to a speed of 12.4 meters per second
in 3 seconds.
He runs at this speed for the next 5 seconds and slows down over the
last 2 seconds as shown in the speed-time graph above.
He crosses the finish line after 10 seconds.
The total distance covered is 100 m.
( a ) Calculate the distanced he runs in the first 8 seconds.
To calculate the distance he runs in the first 8 seconds, we need to find the total distance covered during each of the three segments: acceleration, constant speed, and deceleration.
1. Acceleration:
- Initial speed (u) = 0 m/s
- Final speed (v) = 12.4 m/s
- Time (t) = 3 seconds
Using the equation for distance covered during acceleration:
\[ s = \frac{1}{2}(u + v) \times t \]
\[ s = \frac{1}{2}(0 + 12.4) \times 3 \]
\[ s = \frac{1}{2} \times 12.4 \times 3 \]
\[ s = 6.2 \times 3 = 18.6 \text{ m} \]
The distance covered during acceleration is 18.6 meters.
2. Constant speed:
- Speed = 12.4 m/s
- Time = 5 seconds
Using the formula for distance when speed is constant:
\[ s = \text{speed} \times \text{time} \]
\[ s = 12.4 \times 5 = 62 \text{ m} \]
The distance covered while running at constant speed is 62 meters.
3. Deceleration:
- Initial speed = 12.4 m/s
- Final speed = 0 m/s
- Time = 2 seconds
Using the formula for distance covered during deceleration:
\[ s = \frac{1}{2}(u + v) \times t \]
\[ s = \frac{1}{2}(12.4 + 0) \times 2 \]
\[ s = \frac{1}{2} \times 12.4 \times 2 \]
\[ s = 6.2 \times 2 = 12.4 \text{ m} \]
The distance covered during deceleration is 12.4 meters.
Therefore, the total distance he runs in the first 8 seconds is:
\[ 18.6 + 62 + 12.4 = 93 \text{ m} \]
The athlete runs a total of 93 meters in the first 8 seconds.