Recover planar position from velocity

Try

A synthetic tracking model reports a particle’s horizontal and vertical velocities, rather than its position. At time t seconds, let v(t)=(2t,3−2t) metres per second for 0≤t≤3. Its initial position is (1,−2) metres. Predict whether it returns to its starting height. Does returning to one coordinate imply returning to the starting point?

Read each component separately. Horizontal velocity is nonnegative and grows; vertical velocity begins positive, reaches zero and becomes negative. A pair of rates is not a pair of coordinates. Multiplying the final velocity by elapsed time would assume the velocity stayed constant, which this model explicitly contradicts.

Predict

The vertical direction changes at t=3/2. Predict where the highest point occurs, then decide whether that instant is also when the particle moves most slowly. A zero vertical component does not make the whole velocity zero: the particle can continue moving horizontally.

Before calculating a traveled length, predict whether it is greater than, equal to or smaller than the straight-line separation of the endpoints. A curved excursion cannot be erased from distance merely because its upward and downward coordinate changes cancel.

Experiment

Recover coordinates by integrating each component from the known initial time:

x(t)=1+∫₀ᵗ2u du=1+t²,

y(t)=−2+∫₀ᵗ(3−2u)du=−2+3t−t².

The symbol u is a temporary integration variable. The upper limit t is the requested time. Keep the initial coordinate outside its corresponding integral. Check the two initial values by substituting t=0, then differentiate to recover the supplied velocity.

Time in seconds	Position in metres	Velocity in metres per second
0	(1,−2)	(0,3)
1	(2,0)	(2,1)
3/2	(13/4,1/4)	(3,0)
2	(5,0)	(4,−1)
3	(10,−2)	(6,−3)

Plot these points on equally scaled position axes and label them with their times. Draw a smooth path consistent with the horizontal and vertical signs, not straight segments asserted to be the exact track. Mark the horizontal tangent at the highest point. The coordinate table and formulas specify the path without relying on color or animation.

For a second experiment, change only the starting point to (−4,5). Every coordinate shifts by (−5,7), while every velocity remains unchanged. Explain why the path’s location changes but its traveled length does not.

Understand

In general, if velocity is v(t)=(vₓ(t),vᵧ(t)) and the initial position is r(t₀)=(x₀,y₀), then

r(t)=(x₀+∫ from t₀ to t of vₓ(u)du, y₀+∫ from t₀ to t of vᵧ(u)du).

Continuous component velocities suffice for the fundamental theorem used here. Each integral gives a signed coordinate change. The initial point fixes both constants; knowing only an initial x-coordinate would leave the vertical position undetermined.

In the worked model, the endpoint displacement is (10,−2)−(1,−2)=(9,0), with magnitude 9 metres. Speed, however, is the norm of instantaneous velocity:

s(t)=√[(2t)²+(3−2t)²]=√(8t²−12t+9).

Traveled distance is ∫₀³s(t)dt. This exact integral already specifies a length; an elementary closed form is unnecessary for the distinction being taught. It is greater than 9 because s(t)≥2t, with strict inequality wherever the vertical velocity is nonzero. Integrating gives distance greater than ∫₀³2t dt=9. Do not replace the integral of a norm by the norm of the component integrals.

To find the slowest instant, complete the square: s(t)²=8(t−3/4)²+9/2. Its minimum occurs at t=3/4, before the highest point at t=3/2. Position height and speed answer different questions. More generally, where speed is positive, differentiating its square gives s′=(v·a)/s, where a=v′ is acceleration and the dot product multiplies corresponding components and adds. If speed vanishes, this division cannot be used without another local argument.

Retracing adds another distinction. For r(t)=((t−1)²,0) on [0,2], the endpoints agree, yet the particle travels from x=1 to x=0 and back. Its displacement is zero and its distance is 2. The image is one segment, but the time parameter tells us that segment is traversed twice.

Apply

For each practice case, retain four separate lines: initial point, integrated position, endpoint displacement and speed integral. Include units. If you claim distance equals endpoint separation, justify the required absence of a change in travel direction rather than relying on a drawing.

In the trigonometric case, check both initial coordinates after integration. In the speed-change case, calculate the dot product rather than interpreting positive acceleration in one coordinate as proof that total speed increases. Keep a time-labeled sketch or coordinate description as evidence of the path and any reversal.

Continue

The parametric motion lesson develops length from a supplied position. This investigation reverses the starting information: position is recovered from velocity and an initial point. Next, use the same Cartesian component derivatives to interpret a polar tangent, where the radius itself is not a slope.