Connecting Position, Velocity, and Acceleration of Functions Using Integrals
Introduced| English | Meaning |
|---|---|
| position/pəˈzɪʃn/ | An object's location relative to a chosen reference point or coordinate system. |
| Displacement/dɪˈspleɪsmənt/ | The change in position from an initial point to a final point, including direction. |
Motion, now run through integrals
- Unit 4 went position → velocity → acceleration by differentiating. Integration goes backward.
- Integrate acceleration to recover velocity; integrate velocity to recover position 位置.
- Each step adds up a rate to get the quantity it changes.
- So integrals rebuild the motion story from the rates.
To recover position from velocity you...
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Position $=$ initial $+\int v\,dt$.
Going up the chain
- Velocity from acceleration: $v(t)=v(t_0)+\displaystyle\int_{t_0}^{t} a(s)\,ds$.
- Position from velocity: $s(t)=s(t_0)+\displaystyle\int_{t_0}^{t} v(s)\,ds$.
- Each needs an initial value (the constant of integration in disguise).
- Integrate the rate, then add where you started.
Teacher-created velocity model: signed areas distinguish displacement 位移 and distance.
Recovering position from velocity requires an ____ value to fix the constant.
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Position $=$ initial position $+\int v\,dt$.
Displacement vs. distance
- Displacement over $[a,b]$ is the signed integral of velocity: $\displaystyle\int_a^b v(t)\,dt$ — net change in position.
- Total distance is the integral of speed $|v|$: $\displaystyle\int_a^b |v(t)|\,dt$ — it never subtracts.
- If the object turns around, displacement and distance differ.
- Displacement can be zero after real motion; distance cannot (unless it never moved).
Signed area under velocity
y = t − 2 (velocity)
The signed area under $v(t)$ is displacement; taking absolute value on each piece (split at $v=0$) gives distance.
For $v(t)=t-2$ on $[0,3]$, find the displacement $\int_0^3 (t-2)\,dt$.
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$\tfrac92-6=-\tfrac32$.
Displacement can be zero even when the object has actually moved.
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Out-and-back gives zero net displacement but positive distance.
Distance means splitting at the turns
- To get total distance, find where $v=0$ (the turning points) and integrate $|v|$ piece by piece.
- On each piece $v$ keeps one sign, so $|v|$ is just $\pm v$ there.
- Add the absolute areas of the pieces — no cancellation.
- This "split at $v=0$" step is exactly what makes distance $\neq$ displacement.
For $v(t)=t-2$ on $[0,3]$, total distance is $\int_0^3 |t-2|\,dt$. Compute it.
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Split at $t=2$: $2+\tfrac12=\tfrac52$.
Total distance traveled over $[a,b]$ is...
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Integrate speed $|v|$; splitting at turning points is required.
Displacement ($\int v\,dt$) is signed and can be zero even after moving; total distance ($\int|v|\,dt$) is never negative. To get distance, split the interval where $v=0$ and add the absolute values — do not just integrate $v$ and take its absolute value at the end. And recovering position/velocity needs the initial value.
A particle has $v(t)=t-2$ (m/s) on $[0,3]$. Find displacement and total distance.
- Displacement: $\displaystyle\int_0^3 (t-2)\,dt=\Big[\tfrac{t^2}{2}-2t\Big]_0^3=\tfrac92-6=-\tfrac32$ m.
- $v=0$ at $t=2$: negative on $[0,2)$, positive on $(2,3]$.
- Distance: $\displaystyle\int_0^2 |t-2|\,dt+\int_2^3|t-2|\,dt = 2 + \tfrac12 = \tfrac52$ m.
Integrals rebuild motion: velocity $=$ initial $+\int a\,dt$, position $=$ initial $+\int v\,dt$. Displacement over $[a,b]$ is the signed $\int_a^b v\,dt$; total distance is $\int_a^b|v|\,dt$ — split at the turning points ($v=0$) and add absolute values.
Write from displacement 0 and distance 1
- Velocity is t - 1 on [0, 2].
- The object changes direction at t = 1.
What are the displacement and the total distance?
The areas above and below cancel for displacement and add for distance.
Velocity t - 1 on [0, 2]. The total distance is what?
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The two triangles each have area 0.5.