In one-dimensional motion with non-constant acceleration, the acceleration changes with time, so the motion is no longer described by the constant-acceleration kinematic equations. Instead, the starting point is the position function
\(x(t)\).
Velocity and acceleration then come from derivatives of that position function.
From position to velocity and acceleration
If the position is known as a function of time, then
Velocity.
\[
\begin{aligned}
v(t) &= \frac{dx}{dt}
\end{aligned}
\]
Acceleration.
\[
\begin{aligned}
a(t) &= \frac{d^2x}{dt^2}
\end{aligned}
\]
So at a chosen instant \(t_0\), the calculator needs three quantities:
\(x(t_0)\),
\(v(t_0)\),
and
\(a(t_0)\).
When an exact symbolic derivative is not used, those derivatives can be estimated very accurately by numerical methods.
Central differences
The most important finite-difference formulas in this calculator are the centered first- and second-derivative approximations. If
\(h\)
is a small time step, then
Centered first derivative.
\[
\begin{aligned}
v_h(t_0) &= \frac{x(t_0+h)-x(t_0-h)}{2h}
\end{aligned}
\]
Centered second derivative.
\[
\begin{aligned}
a_h(t_0) &= \frac{x(t_0+h)-2x(t_0)+x(t_0-h)}{h^2}
\end{aligned}
\]
Both of these are second-order accurate, meaning their truncation error is proportional to
\(h^2\)
when the function is smooth.
Richardson extrapolation
To improve the derivative estimates, the calculator also evaluates the same formulas at the smaller step
\(h/2\)
and combines the results by Richardson extrapolation. That cancels the leading
\(h^2\)
error term and gives a fourth-order estimate.
Improved first derivative.
\[
\begin{aligned}
v(t_0) &\approx \frac{4v_{h/2}(t_0)-v_h(t_0)}{3}
\end{aligned}
\]
Improved second derivative.
\[
\begin{aligned}
a(t_0) &\approx \frac{4a_{h/2}(t_0)-a_h(t_0)}{3}
\end{aligned}
\]
This is why the calculator asks for a derivative step in function mode. In numerical-data mode, the data spacing
\(\Delta t\)
plays the same role, and the calculator combines step sizes
\(\Delta t\)
and
\(2\Delta t\)
to build the Richardson-improved result.
Worked prompt example
Use the sample function
\(x(t)=t^3-5t^2+2t+1\)
at
\(t_0=1\).
First evaluate the position:
\[
\begin{aligned}
x(1) &= 1^3 - 5(1)^2 + 2(1) + 1 \\
&= 1 - 5 + 2 + 1 \\
&= -1
\end{aligned}
\]
Now differentiate analytically to see the expected exact values:
\[
\begin{aligned}
v(t) &= \frac{d}{dt}\left(t^3 - 5t^2 + 2t + 1\right) = 3t^2 - 10t + 2 \\
a(t) &= \frac{d}{dt}\left(3t^2 - 10t + 2\right) = 6t - 10
\end{aligned}
\]
Evaluate those at
\(t=1\):
\[
\begin{aligned}
v(1) &= 3(1)^2 - 10(1) + 2 = -5 \\
a(1) &= 6(1) - 10 = -4
\end{aligned}
\]
So the correct output for this sample is
\(x=-1\),
\(v=-5\),
and
\(a=-4\)
in the chosen units.
Why the graph matters
The graph is not just decoration. It shows the relationship between the three quantities:
| Graph |
What it represents |
What to look for |
| \(x(t)\) |
Position vs. time |
Local slope gives velocity |
| \(v(t)\) |
Velocity vs. time |
Local slope would give jerk; sign gives motion direction |
| \(a(t)\) |
Acceleration vs. time |
Shows how the rate of change of velocity varies |
The animation complements the graph by showing a moving object on a track together with velocity and acceleration arrows. This helps you connect the sign and magnitude of the derivatives to an actual motion picture.
Numerical-data mode
When no analytic position function is available, the same ideas can be applied to equally spaced table values. Suppose the table contains
\(x(t_0-2\Delta t)\),
\(x(t_0-\Delta t)\),
\(x(t_0)\),
\(x(t_0+\Delta t)\),
and
\(x(t_0+2\Delta t)\).
Then the calculator forms
Velocity at step \(\Delta t\).
\[
\begin{aligned}
v_{\Delta t}(t_0) &= \frac{x(t_0+\Delta t)-x(t_0-\Delta t)}{2\Delta t}
\end{aligned}
\]
Velocity at step \(2\Delta t\).
\[
\begin{aligned}
v_{2\Delta t}(t_0) &= \frac{x(t_0+2\Delta t)-x(t_0-2\Delta t)}{4\Delta t}
\end{aligned}
\]
Then it combines them:
\[
\begin{aligned}
v(t_0) &\approx \frac{4v_{\Delta t}(t_0)-v_{2\Delta t}(t_0)}{3}
\end{aligned}
\]
The same pattern is used for acceleration:
\[
\begin{aligned}
a_{\Delta t}(t_0) &= \frac{x(t_0+\Delta t)-2x(t_0)+x(t_0-\Delta t)}{(\Delta t)^2} \\
a_{2\Delta t}(t_0) &= \frac{x(t_0+2\Delta t)-2x(t_0)+x(t_0-2\Delta t)}{(2\Delta t)^2} \\
a(t_0) &\approx \frac{4a_{\Delta t}(t_0)-a_{2\Delta t}(t_0)}{3}
\end{aligned}
\]
This is why the table must be equally spaced and why the chosen
\(t_0\)
needs two rows on each side.
Choosing the step size
In function mode, the step
\(h\)
controls how far the calculator samples the function around the evaluation time. If
\(h\)
is too large, the estimate is not local enough. If
\(h\)
is too small, round-off error can start to dominate because nearby values are subtracted. A moderate small value, such as
\(10^{-3}\),
is often a good default for well-behaved functions.
Summary
| Quantity |
Formula |
Meaning |
| Position |
\(x(t_0)\) |
Location at the chosen instant |
| Velocity |
\(v(t_0)=dx/dt\) |
Instantaneous rate of change of position |
| Acceleration |
\(a(t_0)=d^2x/dt^2\) |
Instantaneous rate of change of velocity |
| Centered velocity estimate |
\(\dfrac{x(t_0+h)-x(t_0-h)}{2h}\) |
Second-order numerical derivative |
| Centered acceleration estimate |
\(\dfrac{x(t_0+h)-2x(t_0)+x(t_0-h)}{h^2}\) |
Second-order numerical second derivative |
| Richardson improvement |
\(\dfrac{4D_{h/2}-D_h}{3}\) |
Removes the leading \(h^2\) truncation error |
