Straight-Line Motion: Connecting Position, Velocity, and Acceleration
Introduced| English | Meaning |
|---|---|
| position/pəˈzɪʃn/ | An object's location relative to a chosen reference point or coordinate system. |
| velocity/vəˈlɒsɪti/ | The rate of change of displacement. |
| acceleration/əkˌseləˈreɪʃn/ | The rate of change of velocity. |
| speed/spiːd/ | The distance travelled per unit time. |
| displacement/dɪˈspleɪsmənt/ | The straight-line change in position, including direction. |
Position, velocity, acceleration — a derivative chain
- Motion along a line is the classic home of the derivative.
- Position 位置 $s(t)$ says where the object is at time $t$.
- Its derivative is velocity 速度 $v(t)=s'(t)$ — how fast and which way.
- The derivative of velocity is acceleration 加速度 $a(t)=v'(t)=s''(t)$.
A position-time curve
y = ax³ + bx² + cx
The slope of this position curve is velocity; where it levels off ($v=0$) the object is momentarily at rest.
Velocity is the derivative of...
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$v=s'$: velocity is the derivative of position.
Velocity carries a sign
- Velocity is signed: $v>0$ means moving in the positive direction, $v<0$ the negative direction.
- Speed 速率 is the size only, $|v|$ — always non-negative.
- The object is at rest when $v(t)=0$ (momentarily stopped).
- A sign change in $v$ means the object turned around.
On a velocity-time graph, the area is displacement 位移 and the gradient is acceleration
For $v(t)=3t^2-12t+9$, the object is at rest when $v=0$. One such time is $t=1$; find the other time.
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$3(t^2-4t+3)=3(t-1)(t-3)=0\Rightarrow t=1$ or $t=3$.
Speed is the absolute value of velocity, so speed is never negative.
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Speed $=|v|\ge0$; velocity carries a sign.
Speeding up or slowing down?
- Compare the signs of velocity and acceleration:
- Same sign ($v$ and $a$ both $+$, or both $-$) → the object is speeding up.
- Opposite signs → the object is slowing down.
- (Intuition: acceleration pushing the same way as motion adds speed; pushing against it removes speed.)
An object is speeding up when velocity and acceleration have...
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Same sign → speeding up; opposite → slowing down.
At $t=1$, $v=-2$ and $a=-6$. Select all true statements.
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$v<0$ (negative direction), $v,a$ same sign (speeding up), $|v|=2$. It is not at rest ($v\neq0$).
Reading the whole story
- Displacement over $[t_1,t_2]$ is $s(t_2)-s(t_1)$ — net change in position (can be zero even after moving).
- Direction of motion = sign of $v$; turning points are where $v=0$ and changes sign.
- From $s(t)$ you can get everything by differentiating: $v=s'$, then $a=v'$.
- This chain — position → velocity → acceleration — is one of the most tested ideas in the course.
The net change in position over an interval, $s(t_2)-s(t_1)$, is called the ____.
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Displacement is net change in position.
"Speeding up" is not the same as "positive acceleration." An object with $v<0$ and $a<0$ is speeding up (moving faster in the negative direction). Always compare the signs of $v$ and $a$: same sign → speeding up, opposite signs → slowing down. And velocity (signed) is not speed ($|v|$).
$s(t)=t^3-6t^2+9t$ (metres, seconds). Analyze the motion at $t=1$.
- $v(t)=s'(t)=3t^2-12t+9$; $\;v(1)=3-12+9=0$ — the object is momentarily at rest.
- $a(t)=v'(t)=6t-12$; $\;a(1)=-6$.
- Just after $t=1$, $v$ turns negative while $a<0$ (same sign) → it speeds up moving in the negative direction.
For straight-line motion, velocity $v=s'$ and acceleration $a=v'=s''$. Velocity is signed (direction); speed is $|v|$; the object is at rest when $v=0$. Compare signs: $v$ and $a$ the same → speeding up, opposite → slowing down. Displacement over an interval is the net change $s(t_2)-s(t_1)$.