Sunday, December 30, 2012

Single Variable Calculus, Chapter 2, 2.1, Section 2.1, Problem 6

Suppose that a rock is thrown upward on the planet Mars with a velocity of $10$ m/s, its height in meters $t$ seconds later is given by $10t-1.86t^2$.

a. Find the average velocity over the given time intervals:

(i) [1,2]
(ii) [1,1.5]
(iii) [1,1.1]
(iv) [1,1.01]
(v) [1,1.001]

Average velocity = $\frac{\delta distance}{\delta time} = \frac{y_2 - y_1}{t_2 - t_1}$

$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|c|c|c|c|}
\hline\\
& t_1(s) & t_2(s) & y_1 (m) = 10t_1 - 1.86 t_1^2 & y_2(m) = 10t_2 - 1.86t_2^2 & V_{ave}(m/s) = \frac{y_2 - y_1}{t_2 - t_1} \\
\hline\\
(i) & 1 & 2 & 8.14 & 12.56 & 4.42 \\
\hline\\
(ii) & 1 & 1.5 & 8.14 & 10.815 & 5.35 \\
\hline\\
(iii) & 1 & 1.1 & 8.14 & 8.7494 & 6.094 \\
\hline\\
(iv) & 1 & 1.01 & 8.14 & 8.2026 & 6.26 \\
\hline\\
(v) & 1 & 1.001 & 8.14 & 8.1463 & 6.3\\
\hline
\end{array}

\end{aligned}
\end{equation}
$


b. Estimate the instantaneous velocity when $t=1$

Therefore, the value of instantaneous velocity when $t = 1$ is approximately equal to $6.3$m/s

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