2
$\begingroup$

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.

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ You want to estimate the instantaneous velocity at $t=3$. It would be best to use the points with $t=2$ and $t=4$. The approximation for the instantaneous velocity is just the slope of the line segment connecting the two points (no need to find the equation of the tangent line). Note that slope is just the average velocity of $s$ over $[2,4]$. Better still, would be to take the average of the average velocities over $[2,3]$, $[3,4]$, and $[2,4]$. $\endgroup$ – David Mitra Nov 26 '11 at 11:18
  • $\begingroup$ aaa it means that as you said if we take [2 4] then instantenous velociity is (17.7-5.1)/(4-2)?but it is average velocity $\endgroup$ – dato datuashvili Nov 26 '11 at 11:23
  • $\begingroup$ The instantaneous velocity is $approximately$ the average velocity over $[2,4]$. The best you can do here is to approximate the instantaneous velocity with average velocities. The problem here is to figure out the best way to do that. $\endgroup$ – David Mitra Nov 26 '11 at 11:25
  • $\begingroup$ yes clear but i dont care about best ways now,from tangent equation which i have got,could i calculate it? $\endgroup$ – dato datuashvili Nov 26 '11 at 11:28
  • $\begingroup$ By taking the slope of your tangent line (it's 3.7). But, please observe: $t=1$ and $t=2$ were bad choices to make to form the equation of the line. $\endgroup$ – David Mitra Nov 26 '11 at 11:36
3
$\begingroup$

To answer you directly, you just want the slope of your line: 3.7.

But consider, please:

Below is an accurate scatter plot of your data. Despite what the instructions suggest, you do not know what the graph of $s$ looks like. However, you can imagine a curve that models the data points. This curve is the purple curve shown in the diagram.

enter image description here

Now, the instantaneous velocity at $t=3$ is approximately the slope of the tangent line shown above (approximate because the tangent line shown is tangent to the blue curve and the blue curve approximates the graph of $s$).

How can you estimate this slope using the tabular data? Well, it's essentially what you did: estimate the slope of the tangent line, and hence the instantaneous velocity at $t=3$, with the slope of a secant line between two of the given data points. (Note, please, you only need to estimate the slope of the line; you do not need to find the equation of the tangent line.)

But, you cannot select those two points randomly, this may give a bad estimate. In particular, you want $(3,10.7)$ to be "between" or one of the two points that you choose. Looking at the picture, it should be clear that the best points to choose are $(2,5.1)$ and $(4,17.7)$.

So, we will estimate the instantaneous velocity with the average velocity over $[2,4]$ (the average velocity over $[2,4]$ is the slope of the line connecting the points $[2,5.1]$ and $[4,17.7]$).

$${{\text {inst. vel.}}\atop {\text{ at }} t=3}\approx {17.7-5.1\over 4-2}={12.6\over2}=6.3. $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ and now everything is clear,thanks for answer,i have accepted and also upvoted it $\endgroup$ – dato datuashvili Nov 26 '11 at 12:47
  • $\begingroup$ so it means that when we have some number for example t=5,so i should take [4 6] interval yes?am i right? $\endgroup$ – dato datuashvili Nov 26 '11 at 12:49
  • $\begingroup$ Almost, you were given no information for $t=6$; but maybe $[4,5]$. $\endgroup$ – David Mitra Nov 26 '11 at 12:54
  • $\begingroup$ ok thanks ones again $\endgroup$ – dato datuashvili Nov 26 '11 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.