AcademyOne-Dimensional Motion
Academy
Acceleration as a Rate of Change
Level 1 - Physics topic page in One-Dimensional Motion.
Principle
Acceleration measures how velocity changes with time. In one dimension, its sign tells how velocity changes, not by itself whether speed is increasing.
Notation
\(a_x\)
acceleration on the axis
\(\mathrm{m\,s^{-2}}\)
\(v_x\)
instantaneous velocity
\(\mathrm{m\,s^{-1}}\)
\(\Delta v_x\)
velocity change
\(\mathrm{m\,s^{-1}}\)
\(\Delta t\)
elapsed time
\(\mathrm{s}\)
Method
Velocity change
\[\Delta v_x = v_{xf} - v_{xi}\]
Average rate
\[\bar{a}_x = \frac{\Delta v_x}{\Delta t}\]
Average acceleration spreads the velocity change over the whole interval.
Instant rate
\[a_x = \frac{dv_x}{dt}\]
At one instant, replace the finite interval with a derivative.
Use velocity definition
\[v_x=\frac{dx}{dt}\]
Second derivative
\[a_x=\frac{d^2x}{dt^2}\]
Acceleration is the derivative of the derivative of position.
The graph below is a velocity-time representation. Its slope is acceleration, and the signed area under it would be displacement.
Compare the signs of velocity and acceleration before deciding whether the object is speeding up or slowing down.
Rules
These are the compact results from the method above.
Average acceleration
\[\bar{a}_x = \frac{\Delta v_x}{\Delta t}\]
Instant acceleration
\[a_x = \frac{dv_x}{dt}\]
Second derivative
\[a_x = \frac{d^2x}{dt^2}\]
Examples
Question
Velocity changes from
\[6\,\text{m s}^{-1}\]
to \[-2\,\text{m s}^{-1}\]
in \[4\,\text{s}\]
Find average acceleration.Answer
\[\Delta v_x=-2-6=-8\,\text{m s}^{-1}\]
\[\bar{a}_x=\frac{-8}{4}=-2\,\text{m s}^{-2}\]
Checks
- Acceleration sign follows velocity change.
- Negative acceleration is not always slowing down.
- Speed changes when velocity and acceleration oppose.