Displacement, Velocity, and Acceleration
Đã giới thiệu| English | Ý nghĩa |
|---|---|
| displacement/dɪˈspleɪsmənt/ | Tọa độ |
| velocity/vəˈlɒsɪti/ | vận tốc |
| acceleration/əkˌseləˈreɪʃn/ | gia tốc |
| instantaneous/ˌɪnstənˈteɪnɪəs/ | Tức thời |
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.
On a velocity 速度-time graph the gradient is the acceleration 加速度 and the area is the displacement
You jog once around a $400\ \text{m}$ track and stop where you began. What is your displacement (in m)?
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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 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 ____.
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$\vec{v} = d\vec{x}/dt$ -- velocity is the (vector) rate of change of displacement.
Acceleration
- Acceleration 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.
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Its direction changes, so its velocity changes -- that is acceleration, even at constant speed.
Select all situations that count as accelerating.
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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?
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$v = dx/dt$ 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).
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$v = dx/dt = 4t$, 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.
Write from a velocity of 20
- Position is x = 5 t squared, in metres when t is in seconds.
- Find the velocity at t = 2 s.
Position is x = 5 t squared. What is the velocity at t = 2 s, in m/s?
10 times 2 is 20.
What is the velocity at t = 2 s?
Độ khó 1/5 · 100 xu + một phần thưởng nhỏ cho lần hoàn thành đầu tiên trong ngày
The derivative 10 t is 20 at t = 2. Other options — 12 m/s (the lesson value); 5 m/s (the coefficient left as a speed).