Questions: The displacement (in feet) of a particle moving in a straight line is given by s=(1/2) t^2-9 t+18, where t is measured in seconds. (a) Find the average velocity (in ft / s) over each time interval. (i) [4,8] (ii) [6,8] (iii) [8,10] (iv) [8,12] (b) Find the instantaneous velocity (in ft / s) when t=8.

The displacement (in feet) of a particle moving in a straight line is given by s=(1/2) t^2-9 t+18, where t is measured in seconds.
(a) Find the average velocity (in ft / s) over each time interval.
(i) [4,8]
(ii) [6,8]
(iii) [8,10]
(iv) [8,12]
(b) Find the instantaneous velocity (in ft / s) when t=8.
Transcript text: The displacement (in feet) of a particle moving in a straight line is given by $s=\frac{1}{2} t^{2}-9 t+18$, where $t$ is measured in seconds. (a) Find the average velocity (in $\mathrm{ft} / \mathrm{s}$ ) over each time interval. (i) $[4,8]$ (ii) $[6,8]$ (iii) $[8,10]$ (iv) $[8,12]$ (b) Find the instantaneous velocity (in $\mathrm{ft} / \mathrm{s}$ ) when $t=8$.
failed

Solution

failed
failed

Solution Steps

Step 1: Average Velocity over Interval \([4, 8]\)

The average velocity over the interval \([4, 8]\) is calculated as: \[ \text{Average Velocity} = \frac{s(8) - s(4)}{8 - 4} = -3.0 \, \text{ft/s} \]

Step 2: Average Velocity over Interval \([6, 8]\)

The average velocity over the interval \([6, 8]\) is calculated as: \[ \text{Average Velocity} = \frac{s(8) - s(6)}{8 - 6} = -2.0 \, \text{ft/s} \]

Step 3: Average Velocity over Interval \([8, 10]\)

The average velocity over the interval \([8, 10]\) is calculated as: \[ \text{Average Velocity} = \frac{s(10) - s(8)}{10 - 8} = 0.0 \, \text{ft/s} \]

Step 4: Average Velocity over Interval \([8, 12]\)

The average velocity over the interval \([8, 12]\) is calculated as: \[ \text{Average Velocity} = \frac{s(12) - s(8)}{12 - 8} = 1.0 \, \text{ft/s} \]

Step 5: Instantaneous Velocity at \(t = 8\)

The instantaneous velocity at \(t = 8\) is given by the derivative of the displacement function evaluated at \(t = 8\): \[ \text{Instantaneous Velocity} = s'(8) = -1.0 \, \text{ft/s} \]

Final Answer

The results are summarized as follows:

  • Average velocity over \([4, 8]\): \(-3.0 \, \text{ft/s}\)
  • Average velocity over \([6, 8]\): \(-2.0 \, \text{ft/s}\)
  • Average velocity over \([8, 10]\): \(0.0 \, \text{ft/s}\)
  • Average velocity over \([8, 12]\): \(1.0 \, \text{ft/s}\)
  • Instantaneous velocity at \(t = 8\): \(-1.0 \, \text{ft/s}\)

Thus, the final answers are: \[ \boxed{ \begin{align_} \text{(i)} & \, -3.0 \, \text{ft/s} \\ \text{(ii)} & \, -2.0 \, \text{ft/s} \\ \text{(iii)} & \, 0.0 \, \text{ft/s} \\ \text{(iv)} & \, 1.0 \, \text{ft/s} \\ \text{(b)} & \, -1.0 \, \text{ft/s} \end{align_} }

Was this solution helpful?
failed
Unhelpful
failed
Helpful