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 $t=8$.
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Solution

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Solution Steps

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

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

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

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

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

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

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

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

Step 5: Instantaneous Velocity at t=8t = 8

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

Final Answer

The results are summarized as follows:

  • Average velocity over [4,8][4, 8]: −3.0 ft/s-3.0 \, \text{ft/s}
  • Average velocity over [6,8][6, 8]: −2.0 ft/s-2.0 \, \text{ft/s}
  • Average velocity over [8,10][8, 10]: 0.0 ft/s0.0 \, \text{ft/s}
  • Average velocity over [8,12][8, 12]: 1.0 ft/s1.0 \, \text{ft/s}
  • Instantaneous velocity at t=8t = 8: −1.0 ft/s-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_} }

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