Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
![x = V*t = -8\frac{m}{s} *3s = -24m](https://tex.z-dn.net/?f=x%20%3D%20V%2At%20%3D%20-8%5Cfrac%7Bm%7D%7Bs%7D%20%2A3s%20%3D%20-24m)
After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:
1. The velocity will start to decrease untill it reaches 0m/s.
2. Then, the velocity will start to increase at the rate of the acceleration.
The distance that the ball travels in the first phase can be found with the following expression:
![v^2 = v_0^2 + 2a*d](https://tex.z-dn.net/?f=v%5E2%20%3D%20v_0%5E2%20%2B%202a%2Ad)
Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:
![d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m](https://tex.z-dn.net/?f=d%20%3D%20%5Cfrac%7Bv%5E2%20-%20v_0%5E2%7D%7B2a%7D%20%3D%20%5Cfrac%7B%280m%2Fs%29%5E2%20-%20%28-8m%2Fs%29%5E2%7D%7B2%2A7m%2Fs%5E2%7D%3D%20-4.57m)
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
![t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s](https://tex.z-dn.net/?f=t_1%20%3D%20%5Cfrac%7Bv%7D%7Ba%7D%20%3D%20%5Cfrac%7B8m%2Fs%7D%7B7m%2Fs%5E2%7D%20%3D%201.143%20s)
Then, the time of the second phase will be:
![t_2 = 9s - t_1 = 9s - 1.143s = 7.857s](https://tex.z-dn.net/?f=t_2%20%3D%209s%20-%20t_1%20%3D%209s%20-%201.143s%20%3D%207.857s)
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
![x = \frac{1}{2}a*t^2 + v_0*t + x_0](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7Da%2At%5E2%20%2B%20v_0%2At%20%2B%20x_0)
V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:
![x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A7%5Cfrac%7Bm%7D%7Bs%5E2%7D%2At%5E2%20%2B%200m%2Fs%2At%20%2B%200m%20%3D%20216.07m)
So, the total distance covered by this object in meters will be the sum of all the distances we found:
![x_total = 24m + 4.57m + 216.07m = 244.64m](https://tex.z-dn.net/?f=x_total%20%3D%2024m%20%2B%204.57m%20%2B%20216.07m%20%3D%20244.64m)