Average velocity \( = \frac{-0.25}{0.5} = -0.5 \) m/s
Step-by-step explanation:
To find the average velocity over a given time interval, we'll use the formula:
\[ \text{Average velocity} = \frac{\text{Change in displacement}}{\text{Change in time}} \]
a.
i. Average velocity over \([3,4]\):
First, let's find the displacement at \( t = 3 \) and \( t = 4 \):
- At \( t = 3 \):
\[ s(3) = (3)^2 - 9(3) + 14 = 9 - 27 + 14 = -4 \]
- At \( t = 4 \):
\[ s(4) = (4)^2 - 9(4) + 14 = 16 - 36 + 14 = -6 \]
Now, let's find the change in displacement and change in time:
- Change in displacement: \( -6 - (-4) = -2 \)
- Change in time: \( 4 - 3 = 1 \)
Average velocity \( = \frac{-2}{1} = -2 \) m/s
ii. Average velocity over \([3.5, 4]\):
We can use the same approach as above, evaluating the displacement at \( t = 3.5 \) and \( t = 4 \):
- At \( t = 3.5 \):
\[ s(3.5) = (3.5)^2 - 9(3.5) + 14 = 12.25 - 31.5 + 14 = -5.25 \]
Now, let's find the change in displacement and change in time:
- Change in displacement: \( -6 - (-5.25) = -0.75 \)
- Change in time: \( 4 - 3.5 = 0.5 \)
Average velocity \( = \frac{-0.75}{0.5} = -1.5 \) m/s
iii. Average velocity over \([4,5]\):
The displacement at \( t = 4 \) is already found as \( -6 \), and at \( t = 5 \):
\[ s(5) = (5)^2 - 9(5) + 14 = 25 - 45 + 14 = -6 \]
Change in displacement: \( -6 - (-6) = 0 \)
Change in time: \( 5 - 4 = 1 \)
Average velocity \( = \frac{0}{1} = 0 \) m/s
iv. Average velocity over \([4,4.5]\):
Displacement at \( t = 4.5 \):
\[ s(4.5) = (4.5)^2 - 9(4.5) + 14 = 20.25 - 40.5 + 14 = -6.25 \]
Change in displacement: \( -6.25 - (-6) = -0.25 \)
Change in time: \( 4.5 - 4 = 0.5 \)
Average velocity \( = \frac{-0.25}{0.5} = -0.5 \) m/s
