At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
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.
### 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.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
