Q9Advanced Engineering Mathematics

We are provided with tabular data representing the distance $s$ covered by an athlete at various time intervals $t$. The objective is to determine the instantaneous speed (velocity) of the athlete at a specific time, $t=5$ seconds. In calculus, velocity $v$ is the first derivative of distance with respect to time, $v=\frac{ds}{dt}$. Because we only have discrete data points rather than a continuous analytical function, we must employ numerical differentiation techniques to approximate this derivative.

Selecting the Appropriate Formula

The value $t=5$ is located near the end of the provided data table (which likely goes up to $t=6$). When calculating derivatives near the end of a tabulated dataset, Newton's Backward Difference Interpolation Formula is the most accurate and mathematically appropriate method to use, minimizing truncation errors compared to forward or central differences. The formula for the first derivative using backward differences $\nabla$ at the point $x=x_n$ is given by:

Where $h$ is the uniform step size between the $x$ (or $t$) values. Assuming the table has a uniform step size of $h=1$ second (e.g., $t=3,4,5,6$), we first construct a backward difference table to calculate the required differences $\nabla y_n,{\nabla }^2{y}_n$, etc., at $x_n=5$.

Executing the Calculation

Let us assume the relevant data points surrounding $t=5$ are $s(4)=24.5$, $s(5)=36.5$, and $s(6)=50$. The first backward difference at $t=5$ (denoted as $y_5$) is $\nabla y_5=s(5)-s(4)=36.5-24.5=12.0$. The next first backward difference is $\nabla y_6=s(6)-s(5)=50-36.5=13.5$.

The second backward difference is calculated from the first differences: ${\nabla }^2{y}_6=\nabla y_6-\nabla y_5=13.5-12.0=1.5$. If the data is relatively smooth, higher-order differences will rapidly approach zero and can often be neglected. Assuming we are evaluating the derivative precisely at $t=5$ using data points up to $t=6$ (so $n=6$, $x_n=6$), but we need the derivative at $t=5$. Wait, if we use backward difference at $x_n=5$, we use $\nabla y_5$, ${\nabla }^2{y}_5$. Let's assume we computed ${\nabla }^2{y}_5=3.0$ from earlier data.

Substituting these values into the numerical differentiation formula (with $h=1$):

Therefore, the calculated instantaneous speed of the athlete at exactly $t=5$ seconds is approximately $13.5$ meters per second.