M001 · Kinematics

Motion Graph Studio

Can one drawn curve determine the whole motion? A position curve does: its slope is the velocity, and the way the slope changes is the acceleration. A velocity curve only tells you how far the object moves, so you also need its starting position; an acceleration curve needs the starting velocity too.

Draw the position, velocity or acceleration curve and watch a 3D cart follow it. Slopes and areas, with their signs, show when the cart turns back and why distance and displacement differ.

Selected time $\text{Ready}$

Reference position $\text{Ready}$

Reference signed velocity $\text{Ready}$

Reference signed acceleration $\text{Ready}$

Selected driving quantity $\text{Ready}$

Signed velocity area $\text{Ready}$

Travelled distance $\text{Ready}$

Signed acceleration area $\text{Ready}$

Draw a motion, then scrub and compare

Follow the signs

Driver: 0 position; 1 velocity; 2 acceleration

j

1.00\,

Duration

T

6.00\,s

Integration initial position

x_0

0.00\,m

Integration initial velocity

v_0

0.00\,\frac{\mathrm m}{\mathrm s}

Driver at knot 0

q(0T/6)

0.00\,

Driver at knot 1

q(1T/6)

2.00\,

Driver at knot 2

q(2T/6)

2.00\,

Driver at knot 3

q(3T/6)

0.00\,

Driver at knot 4

q(4T/6)

-2.00\,

Driver at knot 5

q(5T/6)

-2.00\,

Driver at knot 6

q(6T/6)

0.00\,

Replay fraction

\tau

0.40\,

Choose whether you draw position, velocity or acceleration. Seven equally spaced knots define a natural cubic spline for that quantity. Natural means only that its second derivative is zero at the two outer knots; the cart does not have to start or end at rest. Between knots the curve can overshoot the knot values.

In every mode, position, velocity and acceleration all change continuously. Each integral carries its constant across every knot. The initial position is used only when you draw velocity or acceleration, and the initial velocity only when you draw acceleration. When you switch mode, the knot values stay the same and are read in the newly displayed units.

The blue area under the velocity curve is the displacement, with its sign; the copper area under the acceleration curve is the change in velocity, with its sign. Distance travelled comes from cutting the exact position polynomial at every isolated instant where the velocity is zero and adding up the size of each piece. Time spent at rest adds no distance. A separate trapezoid sum over 480 intervals is shown for comparison and keeps its sampling error.

The motion is prescribed directly; no forces are computed. The rail stretches to cover the whole motion, and there are no walls. The cart's size and the raised arrows are schematic. The blue velocity arrow and the copper acceleration arrow each have their own scale, fixed for a given trajectory. All quantitative plots are below the apparatus.

Drag a blue knot or use its slider, which also works from the keyboard. Drag the gold clock to look at the motion up to any moment. The export contains the model curve, not measured data. Nothing plays automatically, and the plots and controls still work without WebGL.

The motion is prescribed for teaching. The Lab states plainly how the curves are interpolated and what the measurements can and cannot show.

Start with the essentials

Focus question: Can one drawn curve determine the whole motion?
One-sentence intuition: Integrating a graph needs a starting value, and area below the axis counts as negative.

Core mathematical model

Velocity and acceleration are slopes

v=\frac{\mathrm dx}{\mathrm dt},\qquad a=\frac{\mathrm dv}{\mathrm dt}

The three curves describe one motion, so editing the driver changes all three together.

Integrals need initial values

\begin{aligned}v(t)&=v_0+\int_0^t a(s)\,\mathrm ds\\x(t)&=x_0+\int_0^t v(s)\,\mathrm ds\end{aligned}

After a reversal, positive and negative areas can cancel.

Distance counts the trip back

D(t)=\int_0^t|v(s)|\,\mathrm ds\geq|x(t)-x(0)|

The zeros of the velocity split each polynomial piece into segments of one-way travel.

Natural boundary of the chosen driver

q\in\{x,v,a\},\qquad q^{\prime\prime}(0)=q^{\prime\prime}(T)=0

The input spline has a continuous second derivative. A natural boundary does not mean zero speed.

Common difficulties

What a natural spline boundary means

Typical misconceptionA natural spline boundary means the cart is at rest at the ends.

Better mental modelYou choose position, velocity or acceleration as the driver, and seven equally spaced knots define a natural cubic spline for it. Only the driver’s second derivative is zero at the two outer ends; the cart does not have to start or end at rest. The spline can also overshoot the knot values.

Run the experiment

01
Select the driver
Compare Draw position, Uniform motion and Constant acceleration.
What to observe: Initial values matter only for curves obtained by integration.
02
Edit and reverse
Drag a blue knot across zero, then drag the gold clock.
What to observe: While the velocity is negative, the cart moves back along the rail.
03
Separate distance and displacement
Choose Reverse and return and drag the clock to the end.
What to observe: The displacement can return to zero even though the cart has traveled some distance.
04
Check the integral
Compare exact position with the separate trapezoid trace.
What to observe: The sampling error of the trapezoid sum stays visible; it is not forced to zero.

Motion Graph Studio