Displacement, Velocity, and Acceleration
Introduced| English |
|---|
| displacement/dɪˈspleɪsmənt/ |
| velocity/vəˈlɒsɪti/ |
| acceleration/əkˌseləˈreɪʃn/ |
| instantaneous/ˌɪnstənˈteɪnɪəs/ |
Walk around the block, end up nowhere
- Walk all the way around a city block and you have covered plenty of ground.
- Yet you finish exactly where you started -- so your net change in position is zero.
- Physics needs to tell these two apart: how far you travelled versus how far you ended up.
- The distinction runs through all of motion.
Displacement versus distance
- Displacement 位移 is the vector from start to finish -- straight-line, with direction.
- Distance is the scalar total path length -- always positive, no direction.
- Around a loop, distance is large but displacement is zero.
You jog once around a $400\ \text{m}$ track and stop where you began. What is your displacement (in m)?
Displacement is start-to-finish. You end where you began, so it is $0$ -- though the distance was $400\ \text{m}$.
Velocity versus speed
- Velocity 速度 is the rate of change of displacement, a vector: $\vec{v} = \dfrac{d\vec{x}}{dt}$.
- Speed is its magnitude -- how fast, with no direction.
- Average velocity uses total displacement; average speed uses total distance.
The vector rate of change of displacement is called the ____.
$\vec{v} = d\vec{x}/dt$ -- velocity is the (vector) rate of change of displacement.
Acceleration
- Acceleration 加速度 is the rate of change of velocity: $\vec{a} = \dfrac{d\vec{v}}{dt}$.
- Speeding up, slowing down, or turning all count as accelerating.
- Its direction need not match the velocity -- braking points it backward.
Motion graphs
Set an initial velocity and acceleration, then read displacement, velocity and acceleration off the graphs.
A car moving at constant speed around a curve is accelerating.
Its direction changes, so its velocity changes -- that is acceleration, even at constant speed.
Select all situations that count as accelerating.
Any change in velocity -- faster, slower, or a change of direction -- is an acceleration.
The calculus chain
- Position, velocity, and acceleration form a derivative chain: $x \xrightarrow{d/dt} v \xrightarrow{d/dt} a$.
- Differentiate to go down the chain; integrate to climb back up.
- Know any one as a function of time and calculus gives you the others.
If position is $x(t)$, how do you get the acceleration?
$v = dx/dt$ and $a = dv/dt$, so $a$ is the second derivative of position.
For $x(t) = 2t^2$, find the velocity at $t = 3$ (in m/s).
$v = dx/dt = 4t$, so $v(3) = 12\ \tfrac{\text{m}}{\text{s}}$.
A particle's position is $x(t) = t^3 - 6t^2 + 9t$ (metres).
- Velocity: $v = \dfrac{dx}{dt} = 3t^2 - 12t + 9$.
- Acceleration: $a = \dfrac{dv}{dt} = 6t - 12$. At $t = 2$, $a = 0$ -- the velocity is momentarily not changing.
An instantaneous 瞬时 velocity and an average velocity are not the same. Average velocity is total displacement over total time; instantaneous velocity is the derivative at one moment. On a curved position graph they can differ a lot.
Displacement (vector, start-to-finish) differs from distance (scalar path length). Velocity $= d\vec{x}/dt$ and acceleration $= d\vec{v}/dt$ form a derivative chain $x \to v \to a$. Turning counts as accelerating, and instantaneous values come from derivatives, not averages.