Q5Advanced Engineering Mathematics I

Question

A slider in a machine moves along a fixed straight rod. Its distance $x$ (cm) along the rod is given below for various values of time $t$ (sec): | $t \Rightarrow$ | 0 | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | |---|---|---|---|---|---|---|---| | $x \Rightarrow$ | 30.28 | 31.43 | 32.98 | 33.54 | 33.97 | 33.48 | 32.13 | Evaluate: i) Velocity for $t = 0.1, 0.5$ and $0.3$ ii) Acceleration for $t = .02, .33$ and $.58$

Answer 1

Velocity is $d x / d t$ (first derivative) and acceleration is $d^{2} x / d t^{2}$ (second derivative).

Since the points $t = 0.1$ and $t = 0.5$ are near the ends of the table, we use Newton's Forward and Backward difference formulas for derivatives.

For $t = 0.3$ (middle of the table), Stirling's central difference formula provides the most accurate derivative.

We construct a difference table of $x$ against $t$ and apply the standard differentiation formulas: $y^{'} = \frac{1}{h} (Δ y_{0} - \frac{1}{2} Δ^{2} y_{0} + \dots)$ .

Q5Advanced Engineering Mathematics I

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