AcademyOne-Dimensional Motion
Academy
Position and Velocity from Calculus
Level 1 - Physics topic page in One-Dimensional Motion.
Principle
Calculus links motion quantities by differentiation and integration. Derivatives describe local rates, while integrals accumulate signed changes over an interval.
Notation
\(x(t)\)
position function
\(\mathrm{m}\)
\(v_x(t)\)
velocity function
\(\mathrm{m\,s^{-1}}\)
\(a_x(t)\)
acceleration function
\(\mathrm{m\,s^{-2}}\)
\(t\)
time
\(\mathrm{s}\)
Method
Velocity is slope
\[v_x(t)=\frac{dx}{dt}\]
Differentiate position to get the local rate of position change.
Acceleration is slope
\[a_x(t)=\frac{dv_x}{dt}\]
Differentiate velocity to get the local rate of velocity change.
Accumulate velocity
\[\Delta x=\int_{t_i}^{t_f}v_x(t)\,dt\]
Signed area under velocity gives displacement.
Accumulate acceleration
\[\Delta v_x=\int_{t_i}^{t_f}a_x(t)\,dt\]
Signed area under acceleration gives velocity change.
The graph below combines the two views: slope turns a position history into velocity, and area under a velocity history gives displacement.
When velocity changes sign, split an integral before calculating distance; a single signed integral gives displacement.
Rules
These are the compact results from the method above.
Velocity from position
\[v_x(t) = \frac{dx}{dt}\]
Acceleration from velocity
\[a_x(t) = \frac{dv_x}{dt}\]
Displacement from velocity
\[\Delta x = \int_{t_i}^{t_f} v_x(t)\,dt\]
Velocity change
\[\Delta v_x = \int_{t_i}^{t_f} a_x(t)\,dt\]
Examples
Question
For \(x(t)=t^3-2t\), find \(v_x(t)\), \(a_x(t)\), and both values at \(t=2\).
Answer
\[v_x=3t^2-2\]
\[a_x=6t\]
At \(t=2\), \[v_x=10\,\text{m s}^{-1}\]
\[a_x=12\,\text{m s}^{-2}\]
Checks
- Derivatives give instantaneous rates.
- Integrals give accumulated change.
- Initial conditions fix constants.
- Graph area has physical units.