I have a table of measurements s(t):

t[s], s[m]

  1. 0, 0
  2. 12.48, 26.4
  3. 18.06, 52.8
  4. 22.32, 79.2

I have calculated all values for $a$ using $$ a=\frac{2s}{t^2} $$ and simply sticking the values into the formula for each measurement.

I have read here that one should instead use the formula $$ s=ut\ +\ \frac{1}{2}at^2, $$ which can be used for the first measurement since $u=0$ (which is then effectively the same formula I used). My problem is that I don't know how to calculate values for $u$ for each measurement and thus calculate $a$ correctly. Is there perhaps another way? Am I doing something wrong?

  • 1
    $\begingroup$ Plot $\dfrac s t$ against $t$ and from the graph you should be able to get $a$ and $u$. $\endgroup$
    – Farcher
    Oct 27, 2017 at 15:37
  • $\begingroup$ The data fits an equation with $u\neq 0$ at $t=0$. You cannot assume $u=0$ at $t=0$. Note that the $u$ in your second equation is a constant (the velocity at$ t=0$). This data can be fit quite simply with a little thought. $\endgroup$ Oct 27, 2017 at 16:36

3 Answers 3


The number 1 rule when analysing experimental data is:

  1. Draw a graph

And the number 2 rule is

  1. If possible draw a straight line graph

Let's obey rule 1 and graph your data. We get:

Graph 1

OK, it looks like a quadratic as expected. But because it's not a straight line graph it's hard to see whether it's a quadratic or not. To get a straight line graph we have to play around with the data. We expect the distance:time equation to be:

$$ s = ut + \tfrac{1}{2}at^2 $$

Suppose we assume $u=0$, then this simplifies to:

$$ s = \tfrac{1}{2}at^2 $$

So if we graph $s$ against $t^2$ we should get a straight line with a gradient of $\tfrac{1}{2}a$. Well, let's do this. The result is:

Graph 2

The red dots are your data, and the blue line is a straight line that I've drawn to give what I think is the best fist to the points. And, well, the blue straight line looks a pretty good fit. The gradient of the line is $0.158$ m/sec$^2$, and this is $\tfrac{1}{2}a$ so we get:

$$ a = 0.316 \,\text{m/sec}^2 $$

But is this really the best fit? Is $u$ really zero? Well we can take our equation:

$$ s = ut + \tfrac{1}{2}at^2 $$

An divide through by $t$ to get:

$$ \frac{s}{t} = u + \tfrac{1}{2}at $$

This tells us that if we graph $s/t$ against $t$ we should get a straight line with a gradient of $\tfrac{1}{2}$ and a $y$ intercept of $u$. OK, let's try it. We get only three points this time because the first point gives $s/t = 0/0$, which we can't do. Anyhow, with three points the graph looks like:

Graph 3

Again the red dots are your data and the blue straight line is my fit. Although it's a long extrapolation back to the $y$ axis it looks to me as if $u$ is definitely not zero. In fact when I measure the gradient and intercept of my straight line I get:

$$\begin{align} a &= 0.291 \,\text{m/sec}^2 \\ u &= 0.30 \,\text{m/sec} \end{align}$$

Finally, lets go back to your original graph and show your data along with the curve we get if we use the values of $a$ and $u$ above. The graph now looks like:|

Graph 4

I'd say that was a pretty good fit!


Assuming that the velocity is $u(t_1)=0$ at starting position $s(t_1)=0$, then the formula you used is true for the first point. Since you don't know if the acceleration is constant in your problem, for each other calculation, you need to compute the starting velocity from previous points. Example:

From 1$\rightarrow$2:

Suppose that $\Delta t_{12}$ is small enough so that acceleration is almost constant during this time: $$ s(t_2)=s(t_1)+u(t_1)t_2+\frac{1}{2}at_2^2\implies26.4=0+0\times12.48+\frac{1}{2}a(12.48)^2\implies a=0.339\>\text{m/s$^2$} $$ Speed is therefore at point 2 $$ u(t_2)=a\Delta t_{12}+u(t_1)\implies u(12.48)=0.339\times 12.48+0=4.231\>\text{m/s} $$ From 2$\rightarrow$3:

Suppose that $\Delta t_{23}$ is small enough so that acceleration is almost constant during this time: $$ s(t_3)=s(t_2)+u(t_2)\Delta t_{23}+\frac{1}{2}a\Delta t_{23}^2\\\implies52.8=26.4+4.231\times(18.06-12.48)+\frac{1}{2}a(18.06-12.48)^2\\\implies a=0.179\>\text{m/s$^2$} $$ And so on and so forth. Notice here how I used formulas related to constant acceleration. So keep in mind that these answer are only approximations.


After review, assuming that $u(t_1)=0$ is false. Indeed, it is much more convenient to assume that a=constant. Then you can fit the data and find that $$ s(t)=0.146t^2+0.293t, $$ which yields a $\pm0.05$ error at max on the four points.


Have a look at the full equation for displacement as a function of time:

$$s(t) = s_0 + v_0t + \frac{1}{2}at^2$$

Here, $s(t)$ is the position at a function of time, $s_0$ is the position at $t = 0$, $v_0$ is the speed at $t = 0$, and $a$ is the (constant) acceleration. In your particular case, $s_0 = 0\,m$, $v_0 = 0\,\frac{m}{s}$.

Your data consist of times and positions measured at those times. Your initial position and velocity at $t = 0$ is zero throughout the motion. Consequently (in this particular exercise), all you are left with is:

$$s(t) = \frac{1}{2}at^2$$

EDIT: I have made the assumption that the object in motion starts from rest. This was not explicitly stated. If you know the initial velocity, you would use that for $v_0$.


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.