Question

The displacement (in feet) of a particle moving in a straight line is given by\(s = \frac{1}{2}{t^2} - 6t + 23\), where\(t\)is measured in seconds.

(a)Find the average velocity over each time interval:

(i) \(\left( {{\bf{4,8}}} \right)\) (ii)\(\left( {{\bf{6,8}}} \right)\) (iii)\(\left( {{\bf{8,10}}} \right)\) (iv)\(\left( {{\bf{8,12}}} \right)\)

(b)Find the instantaneous velocity when\(t = 8\).

(c)Draw the graph of\({\bf{s}}\)as a function of\(t\)and draw the secant lines whose slopes are the average velocities in part (a). Then draw the tangent line whose slope is the instantaneous velocity in part (b).

Short answer

(a)The average velocity of the particle at the time intervals \(\left( {{\bf{4,8}}} \right)\), \(\left( {{\bf{6,8}}} \right)\), \(\left( {{\bf{8,10}}} \right)\), and \(\left( {{\bf{8,12}}} \right)\) are \(0\;{\rm{ft/s}}\), \(1\;{\rm{ft/s}}\), \(3\;{\rm{ft/s}}\), and \(4\;{\rm{ft/s}}\) respectively.

(b)The instantaneous velocity when \(t = 8\) is \(2\;{\rm{ft/s}}\).

(c)

Step-by-step solution

Find the average velocity between the time \(t\) and \(t + h\)

The displacement of a particle is given by \(s = \frac{1}{2}{t^2} - 6t + 23\).

The average velocity can be calculated as:

\(\begin{aligned}{\rm{Average}}\;{\rm{velocity}} &= \frac{{s\left( {t + h} \right) - s\left( t \right)}}{{\left( {t + h} \right) - t}}\\ &= \frac{{\frac{1}{2}{{\left( {t + h} \right)}^2} - 6\left( {t + h} \right) + 23 - \left( {\frac{1}{2}{t^2} - 6t + 23} \right)}}{h}\\ &= \frac{{\frac{1}{2}{t^2} + th + \frac{1}{2}{h^2} - 6t - 6h + 23 - \frac{1}{2}{t^2} - 6t + 23}}{h}\\ &= \frac{{th + \frac{1}{2}{h^2} - 6h}}{h}\\ &= \frac{{h\left( {t + \frac{1}{2}h - 6} \right)}}{h}\\ &= t + \frac{1}{2}h - 6\;{\rm{ft/s}}\end{aligned}\)

(a) Step 2: Find the average velocity over time interval

(i) \(\left( {{\bf{4,8}}} \right)\)

As\(t = 4\)and\(h = 8 - 4 = 4\).

Substitute above values into\({\rm{Average}}\;{\rm{velocity}} = \left( {t + \frac{1}{2}h - 6} \right)\;{\rm{ft/s}}\).

\(\begin{aligned}{\rm{Average}}\;{\rm{velocity}} &= 4 + \frac{1}{2}\left( 4 \right) - 6\\ &= 0\;{\rm{ft/s}}\end{aligned}\)

(ii) \(\left( {{\bf{6,8}}} \right)\)

As\(t = 6\)and\(h = 8 - 6 = 2\).

Substitute above values into\({\rm{Average}}\;{\rm{velocity}} = \left( {t + \frac{1}{2}h - 6} \right)\;{\rm{ft/s}}\).

\(\begin{aligned}{\rm{Average}}\;{\rm{velocity}} &= 6 + \frac{1}{2}\left( 2 \right) - 6\\ &= 1\;{\rm{ft/s}}\end{aligned}\)

(iii) \(\left( {{\bf{8,10}}} \right)\)

As\(t = 8\)and\(h = 10 - 8 = 2\).

Substitute above values into\({\rm{Average}}\;{\rm{velocity}} = \left( {t + \frac{1}{2}h - 6} \right)\;{\rm{ft/s}}\).

\(\begin{aligned}{\rm{Average}}\;{\rm{velocity}} &= 8 + \frac{1}{2}\left( 2 \right) - 6\\ &= 3\;{\rm{ft/s}}\end{aligned}\)

(iv) \(\left( {{\bf{8,12}}} \right)\)

As\(t = 8\)and\(h = 12 - 8 = 4\).

Substitute above values into\({\rm{Average}}\;{\rm{velocity}} = \left( {t + \frac{1}{2}h - 6} \right)\;{\rm{ft/s}}\).

\(\begin{aligned}{\rm{Average}}\;{\rm{velocity}} &= 8 + \frac{1}{2}\left( 4 \right) - 6\\ &= 4\;{\rm{ft/s}}\end{aligned}\)

Thus, the average velocity of the particle at the time intervals \(\left( {{\bf{4,8}}} \right)\), \(\left( {{\bf{6,8}}} \right)\), \(\left( {{\bf{8,10}}} \right)\), and \(\left( {{\bf{8,12}}} \right)\) are \(0\;{\rm{ft/s}}\), \(1\;{\rm{ft/s}}\), \(3\;{\rm{ft/s}}\), and \(4\;{\rm{ft/s}}\) respectively.

(b) Step 3: Use the definition of instantaneous velocity

Apply the definition of instantaneous velocity which states that the position function \(f\left( t \right)\) at time \(t = a\) is represented as \(v\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\).

Find the instantaneous velocity when \(t = 8\)

\(\begin{aligned}v\left( t \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{s\left( {t + h} \right) - s\left( t \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {t + \frac{1}{2}h - 6} \right)\\ &= t - 6\end{aligned}\)

Plug\(t = 8\)into above velocity function.

\(\begin{aligned}v\left( 8 \right) &= 8 - 6\\ &= 2\;{\rm{ft/s}}\end{aligned}\)

Thus, the instantaneous velocity when \(t = 8\) is \(2\;{\rm{ft/s}}\).

(c) Step 5: Draw the graph of \({\bf{s}}\) as a function of \(t\) and draw the secant lines whose slopes are the average velocities in part (a). Then draw the tangent line whose slope is the instantaneous velocity in part (b).