Graph Translation and Terminal Speed

The position reaches a maximum and the drone reverses direction. Position stays continuous because the drone cannot change location instantaneously.

At the zero crossing, the position-time gradient is zero. The gradient then changes sign, so the position has a smooth turning point.

$+7.0\mathrm{m}$ followed by $-11\mathrm{m}$.

From $0$ to $3.0\mathrm{s}$, the areas are $6.0+1.0=7.0\mathrm{m}$. From $3.0$ to $7.0\mathrm{s}$, the signed areas are $-3.0-8.0=-11\mathrm{m}$.

Acceleration is the gradient of the velocity-time graph, while position changes according to its signed area; neither relationship is shape copying.

Use gradient to move from $v$-$t$ to $a$-$t$, and signed area plus initial position to move from $v$-$t$ to $x$-$t$.

From $0$ to $3.0\mathrm{s}$, the drone moves east and slows to rest. From $3.0$ to $7.0\mathrm{s}$, it moves west and speeds up.

Before $3.0\mathrm{s}$, $v>0$ but decreases toward zero. After $3.0\mathrm{s}$, $v<0$ and its magnitude increases.

$-1.0\mathrm{m}{\mathrm{s}}^{-2}$.

$$a=\frac{2.0-4.0}{2.0-0}=-1.0\mathrm{m}{\mathrm{s}}^{-2}.$$

Piecewise constant negative acceleration: $-1.0\mathrm{m}{\mathrm{s}}^{-2}$ from $0$–$2.0\mathrm{s}$, $-2.0\mathrm{m}{\mathrm{s}}^{-2}$ from $2.0$–$3.0\mathrm{s}$, $-1.5\mathrm{m}{\mathrm{s}}^{-2}$ from $3.0$–$5.0\mathrm{s}$, and $-1.0\mathrm{m}{\mathrm{s}}^{-2}$ from $5.0$–$7.0\mathrm{s}$.

Calculate the gradient of each straight velocity-time segment and draw a horizontal acceleration segment for each interval.

The speed and drag are initially very small, so weight dominates and the downward resultant force is close to the weight.

At release, $v=0$, so fluid resistance is negligible in the qualitative model.

As speed increases, drag increases. The resultant force and acceleration decrease to zero when drag balances weight, leaving constant non-zero velocity.

Terminal speed is a force-balance condition: zero resultant force gives zero acceleration, not zero velocity.

Velocity is positive and acceleration is negative. The acceleration opposes the velocity, so the speed decreases.

With downward positive, the still-downward motion has $v>0$. Drag temporarily exceeds weight, so the resultant force and acceleration are upward, $a<0$.

The skydiver cannot change velocity by a finite amount in zero time; the parachute produces a large acceleration over a finite interval instead.

Acceleration is the gradient of velocity. A sharp gradient change is compatible with a continuous velocity curve.

$0\mathrm{m}{\mathrm{s}}^{-2}$.

Constant velocity has zero gradient, so acceleration is zero.

As the falling speed decreases, drag decreases. The difference between drag and weight becomes smaller, so the upward resultant force and acceleration magnitude approach zero.

The forces approach balance again at a smaller constant speed.

The tangent gradient.

Velocity is the instantaneous rate of change of position.

The initial gradient is zero, then the gradient increases and approaches a constant non-zero value.

Rest corresponds to zero position-time gradient. Terminal speed corresponds to a constant non-zero gradient.

The gradient changes rapidly at first and then by progressively smaller amounts; when the gradient becomes constant, acceleration is zero.

Acceleration is the rate of change of velocity, so decreasing curvature indicates decreasing acceleration.

$+18\mathrm{m}{\mathrm{s}}^{-1}$.

Read the ordinate from the first horizontal section.

$0\mathrm{m}{\mathrm{s}}^{-2}$.

The gradient of a horizontal velocity-time segment is zero.

Velocity is positive and acceleration is negative, so the parachutist is still moving downward but slowing.

Opposite signs of velocity and acceleration indicate decreasing speed.

The velocity line has no vertical jump at $t=2.0\mathrm{s}$; only its gradient changes sharply.

The parachute changes the force and acceleration, not the velocity instantaneously.

The velocity-time gradient changes: it is zero, then negative with changing magnitude, then zero again. One SUVAT model requires a single constant acceleration.

Graphical interval reasoning remains valid, but the complete motion does not have one constant acceleration.