Modeling planar motion

A planar-motion model assigns the particle a position $(x(t),\, y(t))$ at each time t. The plotted curve is only the trail: where the particle has been. The motion itself lives in the two coordinate functions, and the same trail can be walked quickly, slowly, or in reverse by different pairs of functions.

So when a question asks "which way is the particle moving at t = 2?", do not stare at the trail's shape. Ask the components. Compare positions at t = 2 and a hair later; whichever way each coordinate moved is the answer, and no elimination or graph is needed.

The division of labor is total:

1. $x(t)$ increasing means the particle moves right; $x(t)$ decreasing means left. The y-component has no vote.
2. $y(t)$ increasing means up; decreasing means down. The x-component has no vote.

Both throttles run at once, so "right and down" or "left and up" are ordinary answers. Example: $x = t^2$, $y = 6 - t$ just after t = 1. x is climbing (1, then 4 by t = 2) while y slides from 5 toward 4: the particle moves right and down. Reading "y is decreasing" as "the particle moves backward" is the classic component mix-up; y only ever speaks about vertical.

A particle reverses horizontal direction exactly where $x(t)$ itself turns around, at a maximum or minimum of the x-component; the same holds vertically for $y(t)$. Take $x = t^2 - 4t$, $y = t$. The x-component is a parabola in t with vertex at t = 2: x falls from 0 down to −4, then climbs back. The particle drifts left until t = 2, reverses, and moves right ever after, all while y carries it steadily upward. The trail is a sideways-opening curve; the reversal is invisible in y.

Two warnings. First, the turnaround is at t = 2, not at t = 4 where x returns to 0: revisiting an old x-value is not reversing. Second, a turnaround needs the component to change direction, not merely to hit zero. The value x = 0 is a place, not an event.

The classic model launches a projectile with $x = v t$ (steady horizontal drift) and $y$ a downward-opening quadratic. Take $x = 30t$, $y = 40t - 5t^2$. The vertical component peaks at t = 4 (the vertex of the quadratic), giving a top height of $y(4) = 160 - 80 = 80$. The flight ends when y returns to 0, at t = 8.

The trap at the top: the particle does not stop there. Only the vertical motion pauses to reverse; the horizontal throttle is still running at a steady 30 per unit of time, so at the peak the particle is moving purely horizontally. "Highest point" is a statement about $y(t)$, and about nothing else.