Answer :
Let's break down the problem step-by-step:
### (a) Determine the position and velocity functions for the coin:
Given the position function for free-falling objects:
[tex]\[ s(t) = -16t^2 + v_0 t + s_0 \][/tex]
For this problem:
- The initial velocity [tex]\( v_0 \)[/tex] is 0 since the coin is dropped.
- The initial position [tex]\( s_0 \)[/tex] is the height of the building, which is 1387 feet.
Thus, the position function simplifies to:
[tex]\[ s(t) = -16t^2 + 1387 \][/tex]
The velocity function [tex]\( v(t) \)[/tex] is the derivative of the position function [tex]\( s(t) \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 1387] \][/tex]
[tex]\[ v(t) = -32t \][/tex]
### (b) Determine the average velocity on the interval [tex]\([3, 4]\)[/tex]:
The average velocity [tex]\( \overline{v} \)[/tex] on an interval [tex]\([t_1, t_2]\)[/tex] is computed as:
[tex]\[ \overline{v} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Given [tex]\( t_1 = 3 \)[/tex] seconds and [tex]\( t_2 = 4 \)[/tex] seconds:
[tex]\[ s(3) = -16(3)^2 + 1387 \][/tex]
[tex]\[ s(3) = -144 + 1387 = 1243 \][/tex]
[tex]\[ s(4) = -16(4)^2 + 1387 \][/tex]
[tex]\[ s(4) = -256 + 1387 = 1131 \][/tex]
Thus, the average velocity is:
[tex]\[ \overline{v} = \frac{1131 - 1243}{4 - 3} = \frac{-112}{1} = -112 \, \text{ft/s} \][/tex]
### (c) Find the instantaneous velocities when [tex]\( t = 3 \)[/tex] seconds and [tex]\( t = 4 \)[/tex] seconds:
The instantaneous velocity at [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ v(3) = -32(3) = -96 \, \text{ft/s} \][/tex]
The instantaneous velocity at [tex]\( t = 4 \)[/tex] seconds:
[tex]\[ v(4) = -32(4) = -128 \, \text{ft/s} \][/tex]
### (d) Find the time required for the coin to reach the ground level:
To find the time [tex]\( t \)[/tex] when the coin reaches the ground level (which corresponds to [tex]\( s(t) = 0 \)[/tex]):
Set the position function to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 1387 \][/tex]
[tex]\[ 16t^2 = 1387 \][/tex]
[tex]\[ t^2 = \frac{1387}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{1387}{16}} \][/tex]
[tex]\[ t = \frac{\sqrt{1387}}{4} \][/tex]
[tex]\[ t \approx 9.179 \, \text{s} \][/tex] (Rounded to three decimal places)
### (e) Find the velocity of the coin at impact:
The velocity at impact is the instantaneous velocity at the time the coin reaches the ground level, [tex]\( t = \frac{\sqrt{1387}}{4} \)[/tex]:
[tex]\[ v\left( \frac{\sqrt{1387}}{4} \right) = -32 \left( \frac{\sqrt{1387}}{4} \right) \][/tex]
[tex]\[ v\left( \frac{\sqrt{1387}}{4} \right) = -8\sqrt{1387} \][/tex]
[tex]\[ v \approx -293.7 \, \text{ft/s} \][/tex] (Rounded to three decimal places)
Thus, the complete solution step-by-step is as follows:
- (a) Position function: [tex]\( s(t) = -16t^2 + 1387 \)[/tex], Velocity function: [tex]\( v(t) = -32t \)[/tex]
- (b) Average velocity on [tex]\([3, 4]\)[/tex]: [tex]\( -112 \, \text{ft/s} \)[/tex]
- (c) Instantaneous velocities: [tex]\( v(3) = -96 \, \text{ft/s} \)[/tex] and [tex]\( v(4) = -128 \, \text{ft/s} \)[/tex]
- (d) Time to reach the ground: [tex]\( t \approx 9.179 \, \text{s} \)[/tex]
- (e) Velocity at impact: [tex]\( -293.7 \, \text{ft/s} \)[/tex]
### (a) Determine the position and velocity functions for the coin:
Given the position function for free-falling objects:
[tex]\[ s(t) = -16t^2 + v_0 t + s_0 \][/tex]
For this problem:
- The initial velocity [tex]\( v_0 \)[/tex] is 0 since the coin is dropped.
- The initial position [tex]\( s_0 \)[/tex] is the height of the building, which is 1387 feet.
Thus, the position function simplifies to:
[tex]\[ s(t) = -16t^2 + 1387 \][/tex]
The velocity function [tex]\( v(t) \)[/tex] is the derivative of the position function [tex]\( s(t) \)[/tex]:
[tex]\[ v(t) = \frac{d}{dt}[-16t^2 + 1387] \][/tex]
[tex]\[ v(t) = -32t \][/tex]
### (b) Determine the average velocity on the interval [tex]\([3, 4]\)[/tex]:
The average velocity [tex]\( \overline{v} \)[/tex] on an interval [tex]\([t_1, t_2]\)[/tex] is computed as:
[tex]\[ \overline{v} = \frac{s(t_2) - s(t_1)}{t_2 - t_1} \][/tex]
Given [tex]\( t_1 = 3 \)[/tex] seconds and [tex]\( t_2 = 4 \)[/tex] seconds:
[tex]\[ s(3) = -16(3)^2 + 1387 \][/tex]
[tex]\[ s(3) = -144 + 1387 = 1243 \][/tex]
[tex]\[ s(4) = -16(4)^2 + 1387 \][/tex]
[tex]\[ s(4) = -256 + 1387 = 1131 \][/tex]
Thus, the average velocity is:
[tex]\[ \overline{v} = \frac{1131 - 1243}{4 - 3} = \frac{-112}{1} = -112 \, \text{ft/s} \][/tex]
### (c) Find the instantaneous velocities when [tex]\( t = 3 \)[/tex] seconds and [tex]\( t = 4 \)[/tex] seconds:
The instantaneous velocity at [tex]\( t = 3 \)[/tex] seconds:
[tex]\[ v(3) = -32(3) = -96 \, \text{ft/s} \][/tex]
The instantaneous velocity at [tex]\( t = 4 \)[/tex] seconds:
[tex]\[ v(4) = -32(4) = -128 \, \text{ft/s} \][/tex]
### (d) Find the time required for the coin to reach the ground level:
To find the time [tex]\( t \)[/tex] when the coin reaches the ground level (which corresponds to [tex]\( s(t) = 0 \)[/tex]):
Set the position function to 0 and solve for [tex]\( t \)[/tex]:
[tex]\[ 0 = -16t^2 + 1387 \][/tex]
[tex]\[ 16t^2 = 1387 \][/tex]
[tex]\[ t^2 = \frac{1387}{16} \][/tex]
[tex]\[ t = \sqrt{\frac{1387}{16}} \][/tex]
[tex]\[ t = \frac{\sqrt{1387}}{4} \][/tex]
[tex]\[ t \approx 9.179 \, \text{s} \][/tex] (Rounded to three decimal places)
### (e) Find the velocity of the coin at impact:
The velocity at impact is the instantaneous velocity at the time the coin reaches the ground level, [tex]\( t = \frac{\sqrt{1387}}{4} \)[/tex]:
[tex]\[ v\left( \frac{\sqrt{1387}}{4} \right) = -32 \left( \frac{\sqrt{1387}}{4} \right) \][/tex]
[tex]\[ v\left( \frac{\sqrt{1387}}{4} \right) = -8\sqrt{1387} \][/tex]
[tex]\[ v \approx -293.7 \, \text{ft/s} \][/tex] (Rounded to three decimal places)
Thus, the complete solution step-by-step is as follows:
- (a) Position function: [tex]\( s(t) = -16t^2 + 1387 \)[/tex], Velocity function: [tex]\( v(t) = -32t \)[/tex]
- (b) Average velocity on [tex]\([3, 4]\)[/tex]: [tex]\( -112 \, \text{ft/s} \)[/tex]
- (c) Instantaneous velocities: [tex]\( v(3) = -96 \, \text{ft/s} \)[/tex] and [tex]\( v(4) = -128 \, \text{ft/s} \)[/tex]
- (d) Time to reach the ground: [tex]\( t \approx 9.179 \, \text{s} \)[/tex]
- (e) Velocity at impact: [tex]\( -293.7 \, \text{ft/s} \)[/tex]
