Final answer:
Using the kinematic equation for uniformly accelerated motion, the soccer ball kicked with an acceleration of 8 m/s² and in the air for 11 seconds would land 484 meters away from where it was kicked.
Step-by-step explanation:
The question is seeking to determine the distance a soccer ball travels when it is kicked and accelerates through the air. Assuming the ball accelerates at a constant rate and that air resistance is negligible, the distance can be found using the kinematic equations for uniformly accelerated motion. The acceleration due to gravity is typically 9.81 m/s² on Earth, but as the acceleration given is 8m/s², this seems to be a modified situation, perhaps a hypothetical or simplified case for the purpose of the question.
If the ball is in the air for 11 seconds and has a horizontal acceleration of 8 m/s², the horizontal distance landed by the ball (d) can be calculated using the formula d = 0.5 × a × t², where a is acceleration and t is the time. Plugging in the values, we get d = 0.5 × 8 m/s² × 11² s² = 484 m. Therefore, the soccer ball would land 484 meters away from the initial point.
In this case, the acceleration is given as 8 m/s² and the time is given as 11 seconds. Plugging in these values, we get:
d = 1/2 * 8 m/s² * (11 s)^2
Simplifying the equation, we get:
d = 1/2 * 8 m/s² * 121 s^2
d = 4 m/s² * 121 s^2
d = 484 m
So the ball landed approximately 484 meters away from where it was kicked.