A motion is given by x = 5t² (metres, seconds). At t = 2
A motion is given by $x = 5t^{2}$ (metres, seconds). At $t = 2\ \text{s}$, compute $\Delta x$ and $dx$ for 1° $\Delta t = 1\ \text{s}$; 2° $\Delta t = 0.1\ \text{s}$; 3° $\Delta t = 0.001\ \text{s}$.
Explanation of Method
Distance travelled: the exact increment against its linear estimate.
Step 1: The exact increment
\[ \Delta x = 5\bigl(2+\Delta t\bigr)^{2}-5\cdot 4 = 5\Bigl[4+4\Delta t+\bigl(\Delta t\bigr)^{2}\Bigr]-20 = 20\,\Delta t+5\bigl(\Delta t\bigr)^{2}. \]
Step 2: The differential
The velocity is $x'_{t} = 10t$, so $x'(2) = 20\ \text{m/s}$: $dx = 20\,\Delta t$.
Physically, $dx$ is the distance that would be covered if the body kept its instantaneous velocity at $t = 2$; the term $5(\Delta t)^{2}$ is the extra distance contributed by the acceleration.
Step 3: The errors in closed form
\[ \bigl|\Delta x-dx\bigr| = 5\bigl(\Delta t\bigr)^{2}, \qquad \frac{\bigl|\Delta x-dx\bigr|}{\bigl|\Delta x\bigr|} = \frac{5(\Delta t)^{2}}{20\Delta t+5(\Delta t)^{2}} = \frac{\Delta t}{4+\Delta t}. \]
Step 4: Case 1°: $\Delta t = 1\ \text{s}$
\[ \Delta x = 20+5 = 25\ \text{m}, \quad dx = 20\ \text{m}, \quad \bigl|\Delta x-dx\bigr| = 5\ \text{m}, \quad \frac{1}{5} = 20\%. \]
Step 5: Case 2°: $\Delta t = 0.1\ \text{s}$
\[ \Delta x = 2+0.05 = 2.05\ \text{m}, \quad dx = 2\ \text{m}, \quad \bigl|\Delta x-dx\bigr| = 0.05\ \text{m}, \quad \frac{0.1}{4.1} \approx 2.44\%. \]
Step 6: Case 3°: $\Delta t = 0.001\ \text{s}$
\[ \Delta x = 0.02+0.000005 = 0.020005\ \text{m}, \quad dx = 0.020\ \text{m}, \quad \bigl|\Delta x-dx\bigr| = 5\cdot10^{-6}\ \text{m}, \quad \frac{0.001}{4.001} \approx 0.025\%. \]
Step 7: Summary
\[\begin{array}{c|c|c|c|c} \Delta t\ (\text{s}) & \Delta x\ (\text{m}) & dx\ (\text{m}) & \text{absolute} & \text{relative} \\ \hline 1 & 25 & 20 & 5 & 20\% \\ 0.1 & 2.050000 & 2.000000 & 0.05 & 2.44\% \\ 0.001 & 0.020005 & 0.020000 & 5\cdot10^{-6} & 0.025\% \end{array}\]
\[\text{Formula check: } \frac{\Delta t}{4+\Delta t} = \frac{1}{5} = 20\%,\ \frac{0.1}{4.1} = 2.439\%,\ \frac{0.001}{4.001} = 0.0250\% \;\;\checkmark\]
\[\text{Always } \Delta x>dx: \text{ the motion accelerates, so the true distance exceeds the constant-velocity estimate} \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\bigl|\Delta x-dx\bigr| = 5\bigl(\Delta t\bigr)^{2}, \qquad \frac{\bigl|\Delta x-dx\bigr|}{\bigl|\Delta x\bigr|} = \frac{\Delta t}{4+\Delta t}\,}\]
Remark
Remark. The discarded term $5(\Delta t)^{2}$ is precisely $\tfrac12 a(\Delta t)^{2}$ with acceleration $a = 10\ \text{m/s}^{2}$ — the familiar second term of the kinematic formula $s = v_{0}t+\tfrac12at^{2}$. So in this example "increment versus differential" is exactly "true displacement versus displacement at constant velocity".