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Answer :
Let's solve this problem step-by-step.
We start with the formula for the height of the pebble above the water surface, given by:
[tex]\[ y = 255 - 16t^2 \][/tex]
Our goal is to find the average velocity for different time intervals and estimate the instantaneous velocity at [tex]\( t = 3 \)[/tex] seconds.
### (a) Average Velocity Calculations
The average velocity over a time interval from [tex]\( t = 3 \)[/tex] to [tex]\( t = 3 + \Delta t \)[/tex] is calculated as:
[tex]\[
\text{Average Velocity} = \frac{y(3 + \Delta t) - y(3)}{\Delta t}
\][/tex]
Where [tex]\( y(t) = 255 - 16t^2 \)[/tex].
For (i) [tex]\(\Delta t = 0.1\)[/tex] seconds:
1. Calculate [tex]\( y(3) \)[/tex]:
[tex]\[
y(3) = 255 - 16(3)^2 = 255 - 144 = 111
\][/tex]
2. Calculate [tex]\( y(3.1) \)[/tex] (since [tex]\( t = 3 + 0.1 = 3.1 \)[/tex]):
[tex]\[
y(3.1) = 255 - 16(3.1)^2 = 255 - 16 \times 9.61 = 255 - 153.76 = 101.24
\][/tex]
3. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{101.24 - 111}{0.1} = \frac{-9.76}{0.1} = -97.6 \, \text{ft/s}
\][/tex]
For (ii) [tex]\(\Delta t = 0.05\)[/tex] seconds:
1. Calculate [tex]\( y(3.05) \)[/tex] (since [tex]\( t = 3 + 0.05 = 3.05 \)[/tex]):
[tex]\[
y(3.05) = 255 - 16(3.05)^2 = 255 - 16 \times 9.3025 = 255 - 148.84 = 106.16
\][/tex]
2. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{106.16 - 111}{0.05} = \frac{-4.84}{0.05} = -96.8 \, \text{ft/s}
\][/tex]
For (iii) [tex]\(\Delta t = 0.01\)[/tex] seconds:
1. Calculate [tex]\( y(3.01) \)[/tex] (since [tex]\( t = 3 + 0.01 = 3.01 \)[/tex]):
[tex]\[
y(3.01) = 255 - 16(3.01)^2 = 255 - 16 \times 9.0601 = 255 - 144.9616 = 110.0384
\][/tex]
2. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{110.0384 - 111}{0.01} = \frac{-0.9616}{0.01} = -96.16 \, \text{ft/s}
\][/tex]
### (b) Instantaneous Velocity Estimation
The instantaneous velocity is the derivative of [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex], evaluated at [tex]\( t = 3 \)[/tex]. The derivative of [tex]\( y = 255 - 16t^2 \)[/tex] is:
[tex]\[
\frac{dy}{dt} = -32t
\][/tex]
Evaluate this at [tex]\( t = 3 \)[/tex]:
[tex]\[
\text{Instantaneous Velocity} = -32 \times 3 = -96 \, \text{ft/s}
\][/tex]
Summary:
- Average Velocities: [tex]\(-97.6 \, \text{ft/s}\)[/tex], [tex]\(-96.8 \, \text{ft/s}\)[/tex], [tex]\(-96.16 \, \text{ft/s}\)[/tex]
- Instantaneous Velocity at [tex]\( t = 3 \)[/tex]: [tex]\(-96 \, \text{ft/s}\)[/tex]
We start with the formula for the height of the pebble above the water surface, given by:
[tex]\[ y = 255 - 16t^2 \][/tex]
Our goal is to find the average velocity for different time intervals and estimate the instantaneous velocity at [tex]\( t = 3 \)[/tex] seconds.
### (a) Average Velocity Calculations
The average velocity over a time interval from [tex]\( t = 3 \)[/tex] to [tex]\( t = 3 + \Delta t \)[/tex] is calculated as:
[tex]\[
\text{Average Velocity} = \frac{y(3 + \Delta t) - y(3)}{\Delta t}
\][/tex]
Where [tex]\( y(t) = 255 - 16t^2 \)[/tex].
For (i) [tex]\(\Delta t = 0.1\)[/tex] seconds:
1. Calculate [tex]\( y(3) \)[/tex]:
[tex]\[
y(3) = 255 - 16(3)^2 = 255 - 144 = 111
\][/tex]
2. Calculate [tex]\( y(3.1) \)[/tex] (since [tex]\( t = 3 + 0.1 = 3.1 \)[/tex]):
[tex]\[
y(3.1) = 255 - 16(3.1)^2 = 255 - 16 \times 9.61 = 255 - 153.76 = 101.24
\][/tex]
3. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{101.24 - 111}{0.1} = \frac{-9.76}{0.1} = -97.6 \, \text{ft/s}
\][/tex]
For (ii) [tex]\(\Delta t = 0.05\)[/tex] seconds:
1. Calculate [tex]\( y(3.05) \)[/tex] (since [tex]\( t = 3 + 0.05 = 3.05 \)[/tex]):
[tex]\[
y(3.05) = 255 - 16(3.05)^2 = 255 - 16 \times 9.3025 = 255 - 148.84 = 106.16
\][/tex]
2. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{106.16 - 111}{0.05} = \frac{-4.84}{0.05} = -96.8 \, \text{ft/s}
\][/tex]
For (iii) [tex]\(\Delta t = 0.01\)[/tex] seconds:
1. Calculate [tex]\( y(3.01) \)[/tex] (since [tex]\( t = 3 + 0.01 = 3.01 \)[/tex]):
[tex]\[
y(3.01) = 255 - 16(3.01)^2 = 255 - 16 \times 9.0601 = 255 - 144.9616 = 110.0384
\][/tex]
2. Determine the average velocity:
[tex]\[
\text{Average Velocity} = \frac{110.0384 - 111}{0.01} = \frac{-0.9616}{0.01} = -96.16 \, \text{ft/s}
\][/tex]
### (b) Instantaneous Velocity Estimation
The instantaneous velocity is the derivative of [tex]\( y \)[/tex] with respect to [tex]\( t \)[/tex], evaluated at [tex]\( t = 3 \)[/tex]. The derivative of [tex]\( y = 255 - 16t^2 \)[/tex] is:
[tex]\[
\frac{dy}{dt} = -32t
\][/tex]
Evaluate this at [tex]\( t = 3 \)[/tex]:
[tex]\[
\text{Instantaneous Velocity} = -32 \times 3 = -96 \, \text{ft/s}
\][/tex]
Summary:
- Average Velocities: [tex]\(-97.6 \, \text{ft/s}\)[/tex], [tex]\(-96.8 \, \text{ft/s}\)[/tex], [tex]\(-96.16 \, \text{ft/s}\)[/tex]
- Instantaneous Velocity at [tex]\( t = 3 \)[/tex]: [tex]\(-96 \, \text{ft/s}\)[/tex]
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