Answer:
To graph the position function f(t), velocity function v(t), and acceleration function a(t) for the given scenario, we'll start by calculating these functions using the given equation:
s(t) = 152t - 16t^2
To find the velocity function, we'll take the derivative of the position function with respect to time:
v(t) = ds(t)/dt = f'(t) = d/dt (152t - 16t^2)
Using the power rule of differentiation, we get:
v(t) = 152 - 32t
Next, to find the acceleration function, we'll take the derivative of the velocity function with respect to time:
a(t) = dv(t)/dt = f''(t) = d/dt (152 - 32t)
Since the derivative of a constant is zero, the acceleration function simplifies to:
a(t) = -32
Now that we have the position function f(t), velocity function v(t), and acceleration function a(t), we can plot their graphs.
(a) Graph of f(t) = 152t - 16t^2:
The graph of f(t) will be a downward-opening parabola with a maximum point. Since the range is restricted to 0 ≤ t ≤ 9.5, the graph will be limited within that range.
(b) Graph of v(t) = 152 - 32t:
The graph of v(t) will be a linear function with a negative slope of -32. It will start with a positive velocity and decrease linearly until it reaches zero at t = 4.75 (half of the time interval), and then it will continue to decrease with negative velocity.
(c) Graph of a(t) = -32:
The graph of a(t) will be a horizontal line at -32, indicating a constant acceleration of -32 ft/s^2.
(d) To complete the statement (e), please provide the specific parts that need to be completed.
(e) Statement incomplete. Please provide the specific parts that need to be completed.
(f) Statement incomplete. Please provide the specific parts that need to be completed.