Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.
