For calculus, I am asked to find instantaneous velocity. Here is the given data and question:
The table shows the position of a cyclist.
$$\begin{array}{c|c|c|c|c|c|c} \hline t\text{ (seconds)}&0 & 1&2&3&4&5\\ \hline s\text{ (meters)} & 0&1.4&5.1&10.7&17.7&25.8\end{array}$$
(a) Find the average velocity for each time period:
$\qquad$ (i) $[1,3]\qquad$ (ii) $[2,3]\qquad$ (iii) $[3,5]\qquad$ (iv) $[3,4]$(b) Use the graph of $s$ as a function of $t$ to estimate the instantaneous velocity when $t=3$.
I think that I should create the equation of tangent line and then put $t=3$ in the equation; for example let's take two arbitrary points, $t=1$ and $t=2$. We have two pairs, $(1,1.4)$ and $(2,5.1)$, and slope $m=\frac{5.1-1.4}{2-1}$ or $m=3.7$, so write equation in slope-form:
$$y-1=3.7(x-1.4)$$
$$y=3.7\cdot x-5.18+1 $$ $$y=3.7\cdot x-4.18$$ First of all, what I wanted to ask was: we have an approximation equation which expresses linear relationship between distance and time, so now I have two choices: directly put 3 into equation, or take the derivative, but derivative couldn't be taken because it is linear so by taking the derivative I will have a constant function, so that means I should put $t=3$, yes?
Please help me.