Answer:
t = (x_ball + 8 ) / 0.6
Step-by-step explanation:
To solve this problem, we can use the equations of motion for the paper ball and Alberta, and then set them equal to each other to find the time and position where they will meet.
The equation of motion for the paper ball is:
x = x0 + v0t + (1/2)at^2
where x is the position of the paper ball, x0 is the initial position (the door), v0 is the initial velocity (2.4 m/s to the right), t is the time, and a is the acceleration (which is 0 for a thrown ball).
The equation of motion for Alberta is:
x = x0 + v0t + (1/2)at^2
where x is the position of Alberta, x0 is the initial position (8 m to the right), v0 is the initial velocity (-0.6 m/s to the left), t is the time, and a is the acceleration (which is 0 for a person running).
If we set these two equations equal to each other, we get:
x_ball = x_alberta
x0 + v0t + (1/2)at^2 = x0 + v0t + (1/2)at^2
This simplifies to:
x_ball = x_alberta
Then we know that both x_ball and x_Alberta are the same point in space (where Alberta met the paper ball)
Now to find time, we need to use one of the original equation for time t. Using equation of motion for Alberta,
x_Alberta = x0 + v0t
So, t = (x_Alberta - x0) / v0
t = (x_ball - 8 ) / (-0.6) = (x_ball + 8 ) / 0.6
The negative sign of velocity of Alberta tells that she is running towards left whereas ball is thrown towards right.
So we can find time and position where Alberta will meet the paper ball by plugging in the specific distance (x_ball) and known values for x0, v0 and acceleration.