Answer :
Let's solve the given problem step-by-step.
The height [tex]\( y \)[/tex] of the pebble above the water surface after [tex]\( t \)[/tex] seconds is given by:
[tex]\[ y = 245 - 16 t^2 \][/tex]
We'll start by addressing each part of the problem.
### (a) Finding the Average Velocity
The average velocity over a time period [tex]\([t_1, t_2]\)[/tex] is given by the formula:
[tex]\[ \text{Average Velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \][/tex]
Where:
- [tex]\( y(t) \)[/tex] is the height of the pebble at time [tex]\( t \)[/tex].
- [tex]\( t_1 \)[/tex] is the initial time.
- [tex]\( t_2 \)[/tex] is the final time.
Given:
- Initial time [tex]\( t_1 = 2 \)[/tex] seconds.
We'll compute the average velocity for the specified durations.
#### (i) For 0.1 seconds
- Final time [tex]\( t_2 = t_1 + 0.1 = 2 + 0.1 = 2.1 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) is:
[tex]\[ y(2) = 245 - 16(2^2) = 245 - 64 = 181 \text{ feet} \][/tex]
- The height at [tex]\( t_2 \)[/tex] (2.1 seconds) is:
[tex]\[ y(2.1) = 245 - 16(2.1^2) = 245 - 16(4.41) = 245 - 70.56 = 174.44 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{174.44 - 181}{2.1 - 2} = \frac{-6.56}{0.1} = -65.6 \text{ ft/s} \][/tex]
#### (ii) For 0.05 seconds
- Final time [tex]\( t_2 = t_1 + 0.05 = 2 + 0.05 = 2.05 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) remains 181 feet (as calculated above).
- The height at [tex]\( t_2 \)[/tex] (2.05 seconds) is:
[tex]\[ y(2.05) = 245 - 16(2.05^2) = 245 - 16(4.2025) = 245 - 67.24 = 177.76 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{177.76 - 181}{2.05 - 2} = \frac{-3.24}{0.05} = -64.8 \text{ ft/s} \][/tex]
#### (iii) For 0.01 seconds
- Final time [tex]\( t_2 = t_1 + 0.01 = 2 + 0.01 = 2.01 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) remains 181 feet (as calculated above).
- The height at [tex]\( t_2 \)[/tex] (2.01 seconds) is:
[tex]\[ y(2.01) = 245 - 16(2.01^2) = 245 - 16(4.0401) = 245 - 64.6416 = 180.3584 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{180.3584 - 181}{2.01 - 2} = \frac{-0.6416}{0.01} = -64.16 \text{ ft/s} \][/tex]
### (b) Estimating the Instantaneous Velocity
The instantaneous velocity of the pebble at a particular time can be found by differentiating the height function [tex]\( y(t) \)[/tex] with respect to [tex]\( t \)[/tex].
Given:
[tex]\[ y = 245 - 16 t^2 \][/tex]
Differentiating with respect to [tex]\( t \)[/tex]:
[tex]\[ v = \frac{dy}{dt} = \frac{d}{dt}(245 - 16 t^2) = -32 t \][/tex]
For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ v = -32 \cdot 2 = -64 \text{ ft/s} \][/tex]
Thus, the instantaneous velocity of the pebble after 2 seconds is:
[tex]\[ v = -64 \text{ ft/s} \][/tex]
### Summary
(a) Average velocities:
(i) -65.6 ft/s
(ii) -64.8 ft/s
(iii) -64.16 ft/s
(b) Instantaneous velocity:
-64 ft/s
The height [tex]\( y \)[/tex] of the pebble above the water surface after [tex]\( t \)[/tex] seconds is given by:
[tex]\[ y = 245 - 16 t^2 \][/tex]
We'll start by addressing each part of the problem.
### (a) Finding the Average Velocity
The average velocity over a time period [tex]\([t_1, t_2]\)[/tex] is given by the formula:
[tex]\[ \text{Average Velocity} = \frac{y(t_2) - y(t_1)}{t_2 - t_1} \][/tex]
Where:
- [tex]\( y(t) \)[/tex] is the height of the pebble at time [tex]\( t \)[/tex].
- [tex]\( t_1 \)[/tex] is the initial time.
- [tex]\( t_2 \)[/tex] is the final time.
Given:
- Initial time [tex]\( t_1 = 2 \)[/tex] seconds.
We'll compute the average velocity for the specified durations.
#### (i) For 0.1 seconds
- Final time [tex]\( t_2 = t_1 + 0.1 = 2 + 0.1 = 2.1 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) is:
[tex]\[ y(2) = 245 - 16(2^2) = 245 - 64 = 181 \text{ feet} \][/tex]
- The height at [tex]\( t_2 \)[/tex] (2.1 seconds) is:
[tex]\[ y(2.1) = 245 - 16(2.1^2) = 245 - 16(4.41) = 245 - 70.56 = 174.44 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{174.44 - 181}{2.1 - 2} = \frac{-6.56}{0.1} = -65.6 \text{ ft/s} \][/tex]
#### (ii) For 0.05 seconds
- Final time [tex]\( t_2 = t_1 + 0.05 = 2 + 0.05 = 2.05 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) remains 181 feet (as calculated above).
- The height at [tex]\( t_2 \)[/tex] (2.05 seconds) is:
[tex]\[ y(2.05) = 245 - 16(2.05^2) = 245 - 16(4.2025) = 245 - 67.24 = 177.76 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{177.76 - 181}{2.05 - 2} = \frac{-3.24}{0.05} = -64.8 \text{ ft/s} \][/tex]
#### (iii) For 0.01 seconds
- Final time [tex]\( t_2 = t_1 + 0.01 = 2 + 0.01 = 2.01 \)[/tex] seconds.
- The height at [tex]\( t_1 \)[/tex] (2 seconds) remains 181 feet (as calculated above).
- The height at [tex]\( t_2 \)[/tex] (2.01 seconds) is:
[tex]\[ y(2.01) = 245 - 16(2.01^2) = 245 - 16(4.0401) = 245 - 64.6416 = 180.3584 \text{ feet} \][/tex]
- The average velocity over this interval is:
[tex]\[ \text{Average Velocity} = \frac{180.3584 - 181}{2.01 - 2} = \frac{-0.6416}{0.01} = -64.16 \text{ ft/s} \][/tex]
### (b) Estimating the Instantaneous Velocity
The instantaneous velocity of the pebble at a particular time can be found by differentiating the height function [tex]\( y(t) \)[/tex] with respect to [tex]\( t \)[/tex].
Given:
[tex]\[ y = 245 - 16 t^2 \][/tex]
Differentiating with respect to [tex]\( t \)[/tex]:
[tex]\[ v = \frac{dy}{dt} = \frac{d}{dt}(245 - 16 t^2) = -32 t \][/tex]
For [tex]\( t = 2 \)[/tex] seconds:
[tex]\[ v = -32 \cdot 2 = -64 \text{ ft/s} \][/tex]
Thus, the instantaneous velocity of the pebble after 2 seconds is:
[tex]\[ v = -64 \text{ ft/s} \][/tex]
### Summary
(a) Average velocities:
(i) -65.6 ft/s
(ii) -64.8 ft/s
(iii) -64.16 ft/s
(b) Instantaneous velocity:
-64 ft/s
