Alright, let’s solve each part of the problem step by step.
### Given:
The height of the ball as a function of time [tex]\( t \)[/tex] in seconds is:
[tex]\[ s(t) = 8t^2 - 3t^3 \][/tex]
### Part (a): Find the velocity and acceleration functions.
To find the velocity, we need to differentiate [tex]\( s(t) \)[/tex] with respect to [tex]\( t \)[/tex].
[tex]\[ v(t) = \frac{d}{dt}[8t^2 - 3t^3] \][/tex]
Calculating the derivative:
[tex]\[ v(t) = \frac{d}{dt} [8t^2] - \frac{d}{dt}[3t^3] \][/tex]
[tex]\[ v(t) = 16t - 9t^2 \][/tex]
To find the acceleration, we need to differentiate the velocity function [tex]\( v(t) \)[/tex] with respect to [tex]\( t \)[/tex].
[tex]\[ a(t) = \frac{d}{dt} [16t - 9t^2] \][/tex]
Calculating the derivative:
[tex]\[ a(t) = 16 - 18t \][/tex]
Thus:
[tex]\[ v(t) = 16t - 9t^2 \][/tex]
[tex]\[ a(t) = 16 - 18t \][/tex]
### Part (b): Over what interval(s) of time is the ball moving in the positive (upward) direction?
The ball moves upwards when the velocity [tex]\( v(t) \)[/tex] is positive:
[tex]\[ v(t) > 0 \][/tex]
So, we solve the inequality:
[tex]\[ 16t - 9t^2 > 0 \][/tex]
Factoring the expression:
[tex]\[ t(16 - 9t) > 0 \][/tex]
Solving for the intervals where the product is positive:
[tex]\[ 0 < t < \frac{16}{9} \][/tex]
Thus, the ball is moving upwards for:
[tex]\[ t \in (0, \frac{16}{9}) \][/tex]
### Part (c): When does the ball return to the ground?
The ball returns to the ground when its height [tex]\( s(t) \)[/tex] is zero:
[tex]\[ s(t) = 0 \][/tex]
So, we solve the equation:
[tex]\[ 8t^2 - 3t^3 = 0 \][/tex]
Factoring out the common term:
[tex]\[ t^2(8 - 3t) = 0 \][/tex]
We get the solutions:
[tex]\[ t = 0 \][/tex]
[tex]\[ 8 - 3t = 0 \][/tex]
[tex]\[ t = \frac{8}{3} = 2.66666666666667 \][/tex] seconds (approximately)
Therefore, the ball returns to the ground at:
[tex]\[ t = 2.667 \][/tex] seconds (rounded to 3 decimal places)
### Part (d): What is the velocity of the ball when it returns to the ground?
To find the velocity when [tex]\( t = 2.667 \)[/tex]:
[tex]\[ v(2.667) = 16(2.667) - 9(2.667)^2 \][/tex]
Using the given result:
[tex]\[ v = -21.333 \][/tex] m/sec (rounded to 3 decimal places)
So, the velocity when the ball returns to the ground is:
[tex]\[ -21.333 \][/tex] m/sec
### Summary:
a. [tex]\[
\begin{array}{l}
v(t) = 16t - 9t^2 \\
a(t) = 16 - 18t
\end{array}
\][/tex]
b. The ball moves upwards for [tex]\( t \in (0, \frac{16}{9}) \)[/tex].
c. The ball returns to the ground at [tex]\( t \approx 2.667 \)[/tex] seconds.
d. The velocity of the ball when it returns to the ground is [tex]\( -21.333 \)[/tex] m/sec.
