Question

If a rock is thrown upward on the Planet Mars with a velocity of 10 m/s, its height in meters t seconds later it is given by \(y = {\bf{10}}t - {\bf{1}}.{\bf{86}}{t^{\bf{2}}}\).

(a) Find the average velocity over the given time intervals:

(i) \(\left( {{\bf{1}},{\bf{2}}} \right)\)

(ii) \(\left( {{\bf{1}},{\bf{1}}.{\bf{5}}} \right)\)

(iii) \(\left( {{\bf{1}},{\bf{1}}.{\bf{1}}} \right)\)

(iv) \(\left( {{\bf{1}},{\bf{1}}.{\bf{01}}} \right)\)

(v) \(\left( {{\bf{1}},{\bf{1}}.{\bf{001}}} \right)\)

(b) Estimate the instantaneous velocity when \(t = {\bf{1}}\).

Short answer

(a) (i) 4.42 m/s (ii) 5.35 m/s (iii) 6.094 m/s (iv) 6.2614 m/s (v) 6.27814 m/s

(b) 6.28 m/s

Step-by-step solution

Step 1:Obtain the height at \(t = {\bf{1}}\)

Substitute 1 for t in the equation \(y = 10t - 1.86{t^2}\).

\(\begin{aligned}{c}y\left( 1 \right) &= 10\left( 1 \right) - 1.86{\left( 1 \right)^2}\\ &= 10 - 1.86\\ &= 8.14\end{aligned}\)

Thus, \(y\left( 1 \right) = 8.14\).

Write the expression for the average velocity

The Step 3: The average velocity in the given intervals between 1 and \(1 + h\) can be written as:

\(\begin{aligned}{c}{v_{avg}} &= \frac{{y\left( {1 + h} \right) - y\left( 1 \right)}}{{\left( {1 + h} \right) - 1}}\\ &= \frac{{\left( {10\left( {1 + h} \right) - 1.86{{\left( {1 + h} \right)}^2}} \right) - 8.14}}{h}\\ &= \frac{{6.28h - 1.86{h^2}}}{h}\end{aligned}\)

Thus, the average velocity is \({v_{avg}} = \frac{{6.28h - 1.86{h^2}}}{h}\).

The average velocity in the given intervals

For the interval \(\left( {1,2} \right)\), and \(h = 1\).

Substitute 1 for h in the equation \({v_{avg}} = \frac{{6.28h - 1.86{h^2}}}{h}\).

\(\begin{aligned}{c}{v_{avg}} &= \frac{{6.28\left( 1 \right) - 1.86{{\left( 1 \right)}^2}}}{1}\\ &= 4.42\;{\rm{m/s}}\end{aligned}\)

For the interval \(\left( {1,1.5} \right)\), and \(h = 0.5\).

Substitute 0.5 for h in the equation \({v_{avg}} = \frac{{6.28h - 1.86{h^2}}}{h}\).

\(\begin{aligned}{c}{v_{avg}} &= \frac{{6.28\left( {0.5} \right) - 1.86{{\left( {0.5} \right)}^2}}}{{0.5}}\\ &= 5.35\;{\rm{m/s}}\end{aligned}\)

For the interval \(\left( {1,1.1} \right)\), and \(h = 0.1\).

Substitute 0.1 for h in the equation \({v_{avg}} = \frac{{6.28h - 1.86{h^2}}}{h}\).

\(\begin{aligned}{c}{v_{avg}} &= \frac{{6.28\left( {0.1} \right) - 1.86{{\left( {0.1} \right)}^2}}}{{0.1}}\\ &= 6.094\;{\rm{m/s}}\end{aligned}\)

For the interval \(\left( {1,1.01} \right)\), and \(h = 0.01\).

Substitute 0.01 for h in the above equation.

\(\begin{aligned}{v_{avg}} &= \frac{{6.28\left( {0.01} \right) - 1.86{{\left( {0.01} \right)}^2}}}{{0.01}}\\ &= 6.2614\;{\rm{m/s}}\end{aligned}\)

For the interval \(\left( {1,1.001} \right)\), \(h = 0.001\).

Substitute 0.001 for h in the equation \({v_{avg}} = \frac{{6.28h - 1.86{h^2}}}{h}\).

\(\begin{aligned}{c}{v_{avg}} &= \frac{{6.28\left( {0.001} \right) - 1.86{{\left( {0.001} \right)}^2}}}{{0.001}}\\ &= 6.27814\;{\rm{m/s}}\end{aligned}\)

The instantaneous velocity

The instantaneous velocity can be calculated by the expression \(\mathop {\lim }\limits_{h \to 0} {v_{avg}}\) as shown below:

\(\begin{aligned}{c}\mathop {\lim }\limits_{h \to 0} {v_{avg}} &= \mathop {\lim }\limits_{h \to 0} \frac{{6.28h - 1.86{h^2}}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \left( {6.28 - 1.86h} \right)\\ &= 6.28\end{aligned}\)

So, the instantaneous velocity is 6.28 m/s.