Numerical Methods for Engineers

Published by McGraw-Hill Science/Engineering/Math
ISBN 10: 0073401064
ISBN 13: 978-0-07340-106-5

Chapter 1 - Mathematical Modeling and Engineering Problem Solving - Problems - Page 24: 1.19b

Answer

\begin{equation} \begin{array}{|c|c|c}{t} & {v(t)} & {x(t)} \\ \hline 0 & {0} & {0} \\ {2} & {19.62} & {0} \\ {4} & {32.0374} & {39.24} \\ {6} & {39.8962} & {103.315} \\ {8} & {44.87} & {183.107} \\ {10} & {48.0179} & {272.847}\end{array} \end{equation}

Work Step by Step

Replace the derivatives of $v(t)$ and $x(t)$ with approximations given by $(1.11)$ to obtain the formula. $$ \begin{aligned} \frac{v\left(t_{i+1}\right)-v\left(t_{i}\right)}{t_{i+1}-t_{i}}=g-\frac{c}{m} v\left(t_{i}\right) & \\ \Rightarrow v\left(t_{i+1}\right)=v\left(t_{i}\right)+\left(g-\frac{c}{m} v\left(t_{i}\right)\right) \cdot\left(t_{i+1}-t_{i}\right) \end{aligned} $$ \begin{array}{l}{\frac{x\left(t_{i+1}\right)-x\left(t_{i}\right)}{t_{i+1}-t_{i}}=v\left(t_{i}\right)} \\ {\Rightarrow x\left(t_{i+1}\right)=x\left(t_{i}\right)+v\left(t_{i}\right) \cdot\left(t_{i+1}-t_{i}\right)}\end{array} Now perform iterations to obtain required approximations __________________________ $v(2)=v(0)+\left(9.81-\frac{12.5}{68.1} v(0)\right) \cdot(2-0)=0+9.81 \cdot 2=19.62$ $x(2)=x(0)+v(0) \cdot(2-0)=0$ $v(4)=19.62+\left(9.81-\frac{12.5}{68.1} \cdot 19.62\right) \cdot 2=32.0374$ $x(4)=0+19.62 \cdot 2=39.24$ $v(6)=32.0374+\left(9.81-\frac{12.5}{68.1} \cdot 32.0371\right) \cdot 2=39.8962$ $x(6)=39.24+39.24 \cdot 2=103.315$ $v(8)=39.8962+\left(9.81-\frac{12.5}{68.1} \cdot 39.8962\right) \cdot 2=44.87$ $x(8)=103.315+39.81-\frac{12.5}{68.1} \cdot 44.87 ) \cdot 2=48.0179$ $v(10)=44.87+\left(9.81-\frac{12.5}{68.1} \cdot 44.87\right) \cdot 2=48.0179$ $x(10)=183.107+44.87 \cdot 2=272.847$
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