I have a homework question I am seeking an alternative solution to. Basically, the question is...
"The table provided below shows the position of a particle S, at several times, t. as the particle moves along a straight line, where t is measured in seconds and S is measured in meters.
$$ \begin{array}{l|cccc} t & 2.0 & 2.7 & 3.2 & 3.8 \\ \hline s(t) & 5.2 & 7.8 & 10.6 & 12.2\\ \end{array} $$
(OR as points (2, 5.2) (2.7, 7.8), (3.2, 10.6) (3.8, 12.2).)
Which of the following best estimates the velocity of the particle at t = 3?
a) 9.2 m/s
b) 7.8 m/s
c) 5.6 m/s"
So what I did was graph the point, drew a tangent line at t = 3 (which approximated to (3, 10.3)). Then I picked another point on the tangent line (2.75, 8.75) that was NOT on the functions path. Then I calculated the slope and got 7.1 m/s, which also gives the velocity. I drew my line EXTREMELY neat with dicimal tick marks. Although I believe this is fine and dandy, I have a strong suspension this is NOT the way the author wanted me to do it.
Could someone offer me some insight on an alternative (maybe algebraic) approach to this question, please?
Thank you in advance
$\endgroup$ 41 Answer
$\begingroup$The best you can do in this case is find the average velocity.
The average velocity, $v_{\text{avg}}$ is given by: $$ v_{\text{avg}}=\frac{\Delta s}{\Delta t} $$ Where $\Delta s$ is the change in displacement and $\Delta t$ is the change in time. ($\Delta$ is the Greek letter "delta" and in this context, it is the final value take away the initial value, so for example $\Delta t=t_{\text{final}}-t_{\text{initial}}$)
At we want the velocity at $t=3$, so we can use the two points around it ($t=2.7\ \text{s}$ and $t=3.2\ \text{s}$): $$ v_{avg}=\frac{\Delta s}{\Delta t}=\frac{10.6-7.8}{3.2-2.7}=5.6\ \text{m}/\text{s} $$
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