The rate θ' at which the baseball player's eyes must move is -4 rad/s.
To find the rate θ' at which the baseball player's eyes must move to watch a fastball with x(t) = -48 m/s as it crosses home plate at x = 0, we can use the following information:
The baseball player stands 2 meters from home plate.
The horizontal distance from the ball to home plate is x.
The angle indicating the direction of the player's gaze is θ.
The fastball is traveling at a speed of 48 m/s.
Since the fastball is traveling at a constant speed, the time it takes for the ball to cross home plate is the time it takes for the ball to travel from the player's position to home plate. Therefore, we can use the time equation for a linear motion:
t = x/v
Substituting the given values, we get:
t = 2/-48 = -1/24
This means that it takes -1/24 seconds for the ball to cross home plate. To find the rate θ' at which the player's eyes must move, we can use the fact that the ball is traveling at a constant speed and the time it takes for the ball to cross home plate is constant. Therefore, the rate of change of the angle θ with respect to time t is:
θ' = dθ/dt = 1/dt
Substituting the time value we found, we get:
θ' = 1/-1/24= -4
Thus, the rate θ' at which the baseball player's eyes must move is -4 rad/s.