Answered

Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Experience the ease of finding reliable answers to your questions from a vast community of knowledgeable experts. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

Help me solve the image provided below

Help Me Solve The Image Provided Below class=

Sagot :

rvkacademic

Answer:

All figures rounded to 4 decimal places

[tex][10, 11]: \quad \boxed{ 10.57 \; m/s}[/tex]

[tex][10, 10.5]: \quad \boxed{ 10.985\; m/s}[/tex]

[tex][10, 10.1]: \quad \boxed{ 11.317\;m/s}[/tex]

[tex][10, 10.01]: \quad \boxed{ 11.3917\;m/s}[/tex]

[tex][10, 10.001]: \quad \boxed{ 11.3992\;m/s}[/tex]

Step-by-step explanation:

The given function for height h in meters in terms of time t in seconds is:

[tex]h(t) = 28t- 0.83t^2[/tex]

To compute the average velocity for each of the intervals provided in the question

  • Find the height of the arrow at the lower and upper limits of the interval
  • Find the difference between these two values
  • Divide the difference by the time difference (the interval length)

Interval [10, 11]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(11) = 28(11) - 0.83 \cdot 11^2 = 207.57 $ meters\\\\Change in height = 207.57 - 197 = 10.57 $ meters\\\\\text{Elapsed time} = 11 - 10 = 1 $ second\\\\\text{Average Velocity} = \dfrac{10.57 \; meters}{1 \;second} = 10.57 $ m/s[/tex]

Interval [10, 10.5]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(10.5) = 28(10.5) - 0.83 \cdot (10.5)^2 = 202.4925 $ meters\\\\Change in height = 202.4925 - 197 = 5.4925 $ meters\\\text{Elapsed time} = 10.5 - 10 = 0.5 $ second\\\\\text{Average Velocity} = \dfrac{ 5.4925 \; meters}{0.5 \;second} = 10.985$ m/s[/tex]

Interval [10, 10.1]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\h(10.1) = 28(10.1) - 0.83\cdot (10.1)^2 = 198.1317 $ meters\\\\\text{Change in height} = 198.1317 - 197 = 1.1317 $meters\\\\\text{Elapsed time} = 10.1 - 10 = 0.1 $ second\\\\\text{Average Velocity} = \dfrac{ 1.1317 \; meters}{0.1 \;second} = 11.317 $ m/s[/tex]

Interval [10, 10.01]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\[/tex]

[tex]h(10.01) = 28(10.01) - 0.83 \cdot (10.01)^2 = 197.113917 $ meters\\[/tex]

[tex]\text{Change in height} = 197.113917 - 197 = 0.113917 $ meters[/tex]

[tex]\text{Elapsed time} = 10.01 - 10 = 0.01 $ second\\[/tex]

[tex]\text{Average Velocity} = \dfrac{ 0.113917\; meters}{0.01 \;second} = 11.3917 $ m/s[/tex]

Interval [10, 10.001]

[tex]h(10) = 28(10) - 0.83\cdot 10^2 = 197 $ meters\\\\[/tex]

[tex]h(10.001) = 28(10.001) - 0.83 \cdot (10.001)^2 = 197.01139917 $ meters\\[/tex]

[tex]\text{Change in height} = 197.01139917 - 197 = 0.01139917 $ meters\\[/tex]

[tex]\text{Elapsed time} = 10.001 - 10 = 0.001 $ second\\[/tex]

[tex]\text{Average Velocity} = \dfrac{ 0.01139917 \; meters}{0.001 \;second} = 11.39917 $ m/s[/tex]

Just a note:
You really don't have to compute h(10) each time, since the lower limit on the interval is always 10. So you can compute h(10) once and this is equal to 197

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.