Finding value of g given a motion map

[kuruman]

There are two possible answers:

1) The wrong graph was included. And the graph shows an object falling from rest at $0.5 cm$ under the Earth's gravity for $0.4s$.

I vote for Answer (1). I think that Answer (2) is untenable. Here is why.

Under constant acceleration, the instantaneous velocity is equal to the average velocity at the middle of the time interval over which the average is taken. Let $t_1$ be the clock time corresponding to event 1. The clock time corresponding to the subsequent events is $t_n=t_1+(n-1)\Delta t, ~n=2,3,\dots$
Now we can compute instantaneous velocities and times at which they occur.
Event 1: $s_1\rightarrow s_2$
$\bar v_1=\dfrac{s_2-s_1}{t_2-t_1}=\dfrac{s_2-s_1}{\Delta t}$
The instantaneous velocity $v_1=\dfrac{s_2-s_1}{\Delta t}$ occurs at clock time $\tau_1=t_1+\frac{1}{2}\Delta t.$

Event 2: $s_3\rightarrow s_2$
$\bar v_2=\dfrac{s_3-s_3}{t_3-t_2}=\dfrac{s_3-s_2}{\Delta t}$
The instantaneous velocity $v_2=\dfrac{s_3-s_2}{\Delta t}$ occurs at clock time $\tau_2=t_2+\frac{1}{2}\Delta t=(t_1+\Delta t)+\frac{1}{2}\Delta t=t_1+\frac{3}{2}\Delta t$

Now we can compute the constant acceleration $$\begin{align} & a_{\text{3-2-1}}=\frac{v_2-v_1}{\tau_2-\tau_1}=\frac{\frac{s_3-s_2}{\Delta t}-\frac{s_2-s_1}{\Delta t}}{t_1+\frac{3}{2}\Delta t-(t_1+\frac{1}{2}\Delta t)} =\frac{s_3-2s_2+s_1}{(\Delta t)^2} \nonumber \\& =\frac{(0.25-2\times 0.1+0.05)\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m/s}^2.
\nonumber \end{align}$$ One needs at least three consecutive positions and times and here we have five. Using the rest of the triplets, $$\begin{align} & a_{\text{4-3-2}}=\frac{s_4-2s_3+s_2}{(\Delta t)^2}=\frac{(0.50-2\times 0.25+0.10)\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m/s}^2 \nonumber \\
& a_{\text{5-4-3}}=\frac{s_5-2s_4+s_3}{(\Delta t)^2}=\frac{(0.85-2\times 0.50+0.25)\rm{m}}{(0.1~\rm{s})^2}=10~\text{m/s}^2 \nonumber \\
\end{align}$$This shows that one can calculate the constant acceleration from three position-time pairs without assuming anything about the initial conditions. Three points are sufficient to define a parabola when there is no experimental uncertainty as in this case.

In my opinion, all this is gorilla-glue strong evidence that the data support an acceleration $a=10~\text{m/s}^2$. In fact, the "triplet analysis" was (and maybe still is) how spark timer data were processed for finding $g$ experimentally before spreadsheets, photogates and lab analysis software came into being. Thus, I am forced to conclude that the graph is wrong. Furthermore, since all accelerations are exactly equal, it is highly likely that we have here a contrived set of data with zero uncertainty.

This concludes my justification of my vote for A. Now let's look at model B:

2) The graph shows an object falling from $0 cm$ for $0.5s$. The question required the answer which best fits the data (not necessarily perfectly fits the data). And, allowing for experimental error, $g = 6.8 m/s^2$ is a good fit for the data.

We can use the model to calculate the positions at the "measured" times. These are summarized in the Table below.

Time (s)​	Expt. Position (m)​	Calc. Position (m)​
0​	N.A.​	0​
0.1​	0.05​	0.03​
0.2​	0.10​	0.14​
0.3​	0.25​	0.30​
0.4​	0.50​	0.54​
0.5​	0.85​	0.85​

The discrepancy between the "experimental" values and the calculated values has been attributed to "experimental" error. I doubt that there ever was an experiment that generated these data, but I will play along. I will superimpose on OP's figure the predictions of the model A where the acceleration is $a_A=10~\text{m/s}^2$ and the object starts from rest at the 0.05 m mark (red circles) and model B where the position are calculated using an acceleration $a_B=6.8~\text{m/s}^2$ and the object starts from rest at the 0 m mark (blue circles.)

A more conventional presentation of model B is shown below. The "experimental" data are shown as discrete points connected with a smoothed line and the calculation is shown as a solid line. It looks like the assumption that the initial position of the object is zero throws the parabola out of kilter. This is no random variation due to experimental uncertainty.

Let the reader decide which is the best analysis of the data: one that yields a value for the constant acceleration that makes assumptions about the initial conditions or one that does not need such assumptions.