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Sagot :
To address the problem, let's go through the required steps one by one:
### How long does it take the rock to hit the ground?
The rock's position as a function of time is given by:
[tex]\[ s(t) = -16t^2 + 230 \][/tex]
The rock hits the ground when [tex]\( s(t) = 0 \)[/tex]. Therefore, we need to solve the equation:
[tex]\[ -16t^2 + 230 = 0 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ -16t^2 = -230 \][/tex]
[tex]\[ t^2 = \frac{230}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{230}{16}} \][/tex]
[tex]\[ t \approx 3.791 \][/tex]
So, it takes approximately [tex]\( \boxed{3.791} \)[/tex] seconds for the rock to hit the ground.
### Find the average velocity of the rock from when it is released until when it hits the ground
The average velocity (v_avg) is defined as the total displacement divided by the total time.
At [tex]\( t = 0 \)[/tex], the position is:
[tex]\[ s(0) = -16(0)^2 + 230 = 230 \text{ ft} \][/tex]
At [tex]\( t = 3.791 \)[/tex], the position is:
[tex]\[ s(3.791) = -16(3.791)^2 + 230 \approx 0 \text{ ft} \][/tex]
The total displacement is:
[tex]\[ s_{\text{final}} - s_{\text{initial}} = 0 - 230 = -230 \text{ ft} \][/tex]
The total time is [tex]\( 3.791 \)[/tex] seconds. Therefore, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{-230 \text{ ft}}{3.791 \text{ s}} \approx -60.663 \text{ ft/s} \][/tex]
So, the average velocity is approximately [tex]\( \boxed{-60.663} \)[/tex] feet per second.
### What time after the rock is thrown will its instantaneous velocity be equal to its average velocity?
The instantaneous velocity [tex]\( v(t) \)[/tex] is the derivative of the position function [tex]\( s(t) \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}(-16t^2 + 230) = -32t \][/tex]
We need to find the time [tex]\( t \)[/tex] when the instantaneous velocity is equal to the average velocity. Set:
[tex]\[ -32t = -60.663 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ t = \frac{60.663}{32} \approx 1.896 \][/tex]
So, the time when the instantaneous velocity is equal to the average velocity is approximately [tex]\( \boxed{1.896} \)[/tex] seconds after it is thrown.
In summary:
- Time to hit the ground: [tex]\( \approx 3.791 \)[/tex] seconds
- Average velocity: [tex]\( \approx -60.663 \)[/tex] feet per second
- Time for instantaneous velocity to match average velocity: [tex]\( \approx 1.896 \)[/tex] seconds
### How long does it take the rock to hit the ground?
The rock's position as a function of time is given by:
[tex]\[ s(t) = -16t^2 + 230 \][/tex]
The rock hits the ground when [tex]\( s(t) = 0 \)[/tex]. Therefore, we need to solve the equation:
[tex]\[ -16t^2 + 230 = 0 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ -16t^2 = -230 \][/tex]
[tex]\[ t^2 = \frac{230}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{230}{16}} \][/tex]
[tex]\[ t \approx 3.791 \][/tex]
So, it takes approximately [tex]\( \boxed{3.791} \)[/tex] seconds for the rock to hit the ground.
### Find the average velocity of the rock from when it is released until when it hits the ground
The average velocity (v_avg) is defined as the total displacement divided by the total time.
At [tex]\( t = 0 \)[/tex], the position is:
[tex]\[ s(0) = -16(0)^2 + 230 = 230 \text{ ft} \][/tex]
At [tex]\( t = 3.791 \)[/tex], the position is:
[tex]\[ s(3.791) = -16(3.791)^2 + 230 \approx 0 \text{ ft} \][/tex]
The total displacement is:
[tex]\[ s_{\text{final}} - s_{\text{initial}} = 0 - 230 = -230 \text{ ft} \][/tex]
The total time is [tex]\( 3.791 \)[/tex] seconds. Therefore, the average velocity is:
[tex]\[ v_{\text{avg}} = \frac{\Delta s}{\Delta t} = \frac{-230 \text{ ft}}{3.791 \text{ s}} \approx -60.663 \text{ ft/s} \][/tex]
So, the average velocity is approximately [tex]\( \boxed{-60.663} \)[/tex] feet per second.
### What time after the rock is thrown will its instantaneous velocity be equal to its average velocity?
The instantaneous velocity [tex]\( v(t) \)[/tex] is the derivative of the position function [tex]\( s(t) \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}(-16t^2 + 230) = -32t \][/tex]
We need to find the time [tex]\( t \)[/tex] when the instantaneous velocity is equal to the average velocity. Set:
[tex]\[ -32t = -60.663 \][/tex]
Solving for [tex]\( t \)[/tex], we get:
[tex]\[ t = \frac{60.663}{32} \approx 1.896 \][/tex]
So, the time when the instantaneous velocity is equal to the average velocity is approximately [tex]\( \boxed{1.896} \)[/tex] seconds after it is thrown.
In summary:
- Time to hit the ground: [tex]\( \approx 3.791 \)[/tex] seconds
- Average velocity: [tex]\( \approx -60.663 \)[/tex] feet per second
- Time for instantaneous velocity to match average velocity: [tex]\( \approx 1.896 \)[/tex] seconds
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