Answer:
244.64m
Explanation:
First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:
[tex]x = V*t = -8\frac{m}{s} *3s = -24m[/tex]
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:
[tex]v^2 = v_0^2 + 2a*d[/tex]
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:
[tex]d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m[/tex]
Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:
[tex]t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s[/tex]
Then, the time of the second phase will be:
[tex]t_2 = 9s - t_1 = 9s - 1.143s = 7.857s[/tex]
Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:
[tex]x = \frac{1}{2}a*t^2 + v_0*t + x_0[/tex]
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:
[tex]x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m[/tex]
So, the total distance covered by this object in meters will be the sum of all the distances we found:
[tex]x_total = 24m + 4.57m + 216.07m = 244.64m[/tex]
