foundation25 min studyLesson 2 in Physics

Kinematics: reading a trajectory

Connect position, velocity, and acceleration while keeping coordinate choices distinct from physical motion.

Before you begin

On this page

The question for today

What you’ll understand

Translate between a trajectory and its derivatives.
Separate displacement from distance and velocity from speed.
Solve a constant-acceleration trajectory with explicit initial data.

A video of a thrown ball gives a sequence of positions. Kinematics turns that sequence into a precise description of motion without yet asking what interaction caused it. The central object is a trajectory: position as a function of time. Once a coordinate system and initial conditions are fixed, derivatives reveal how that motion changes.

Choose coordinates before reading motion

An origin is a reference point, not a privileged place in nature. Moving the origin changes the numbers assigned to every position but leaves displacements unchanged. Reversing an axis changes component signs while preserving the physical path. Draw the axes and state the time origin before assigning values; doing so prevents many apparent disagreements between correct calculations.

Displacement compares final and initial positions. Distance accumulates the length traveled along the entire path. A runner returning to the start has zero displacement and positive distance. Correspondingly, average velocity uses displacement, while average speed uses distance. At an instant, speed is the magnitude of the velocity vector; it discards directional information needed for predicting the next position.

\mathbf v(t)=\frac{d\mathbf r}{dt},\qquad \mathbf a(t)=\frac{d\mathbf v}{dt}

For a differentiable trajectory in a fixed Cartesian frame, velocity has units m/s and acceleration m/s². Nonzero acceleration can change direction even when speed stays constant.↔ Scroll wide equations horizontally.

Slopes and accumulated changes tell one story

The slope of a position–time graph is velocity, not position. A horizontal position graph means rest; a horizontal velocity graph means constant velocity, which may be nonzero. On a velocity–time graph, signed area measures displacement. Negative areas cancel positive ones for displacement, whereas distance requires integrating the absolute speed. Read the axis labels before interpreting any curve.

An acceleration graph alone does not fix a trajectory. Integrating once introduces the initial velocity, and integrating again introduces the initial position. These are physical data, not algebraic annoyances. Two balls can have identical gravitational acceleration but different trajectories because one was dropped and the other launched upward. A differential equation becomes predictive only when the appropriate initial data are supplied.

\mathbf v(t)=\mathbf v_0+\mathbf a t,\qquad \mathbf r(t)=\mathbf r_0+\mathbf v_0t+\frac12\mathbf a t^2

These expressions assume constant acceleration, an inertial Cartesian frame, and initial data at t = 0. They should not be reused unchanged when drag makes acceleration depend on velocity.↔ Scroll wide equations horizontally.

One trajectory can contain several independent components

For a projectile near Earth's surface, neglect air resistance and take gravitational acceleration as uniform and downward. Horizontal acceleration then vanishes, while vertical acceleration is negative if upward is positive. The components share the same time; they are not separate journeys. Solving the vertical equation for a landing time lets you evaluate horizontal position at that very instant.

A turning point illustrates why motion cannot be summarized by speed alone. At the top of a vertical toss, vertical velocity is momentarily zero but acceleration remains downward. A curved trajectory can also maintain constant speed while its velocity direction changes. Later, field theory replaces one particle's trajectory with a value at every spacetime point, but derivative and initial-data reasoning survive that enlargement.

x=x_0+v_{0x}t,\qquad y=y_0+v_{0y}t-\frac12gt^2

The projectile approximation uses g > 0 in m/s², x horizontal, y upward, and ignores drag, Earth's curvature, and changes of g with altitude.↔ Scroll wide equations horizontally.

Work it through

Launch from a low platform

A ball leaves a $1.25\,\mathrm{m}$ platform horizontally at $4.0\,\mathrm{m\,s^{-1}}$ . Use $g=10\,\mathrm{m\,s^{-2}}$ and neglect drag. Find the time to the floor and horizontal range.

Set the floor to $y=0$ , launch time to $t=0$ , and upward positive. Initial data are $y_0=1.25\,\mathrm{m}$ and $v_{0y}=0$ ; therefore $0=1.25-5t^2$ .
The equation gives $t^2=0.25\,\mathrm{s^2}$ . Select $t=0.50\,\mathrm{s}$ because the negative root describes an extrapolation before launch, outside the event being modeled.
Horizontal velocity stays $4.0\,\mathrm{m\,s^{-1}}$ , so $\Delta x=v_{0x}t=2.0\,\mathrm{m}$ . As a check, impact vertical velocity is $-gt=-5.0\,\mathrm{m\,s^{-1}}$ , not zero.

The result

The ball reaches the floor after $0.50\,\mathrm{s}$ , $2.0\,\mathrm{m}$ horizontally from the platform edge. Both answers depend on the idealized no-drag model.

The living notebook

Explore the model

Interactive lab

Interactive lab · vectors

A moving vector still has components

Change the angle and magnitude of one normalized vector and inspect its x/y projections. Pause to describe how direction changes the components. This supports velocity-component reasoning but does not integrate a projectile trajectory.

Vectorx / y projectionsTrace marker

x component: 1.393
y component: 0.975
Magnitude: 1.7

Angle35°Magnitude1.7

t = 0 s

Playback traces a fixed vector and its projections. Angle and magnitude set the geometry; the moving marker does not change them. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA changing direction changes a vector even when its magnitude stays fixed.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

The vertical velocity changes sign through zero at the top. Gravity still acts there, so acceleration remains downward. The same conclusion follows from the derivative of velocity, not from whether the ball is instantaneously moving.

Your turn

Try it yourself

A particle follows $x(t)=2+3t-t^2$ in metres with t in seconds. Find its velocity and acceleration at $t=2\,\mathrm{s}$ and decide whether it is speeding up then.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate once and twice, then compare the signs of velocity and acceleration.

Compare with the solution

Differentiation gives $v=3-2t$ in m/s and $a=-2\,\mathrm{m\,s^{-2}}$ . At two seconds, $v=-1\,\mathrm{m\,s^{-1}}$ . Velocity and acceleration have the same negative direction, so the magnitude of velocity is increasing: the particle is speeding up while moving toward decreasing x.

Keep the idea

Take it with you

A trajectory needs a frame, a clock origin, and initial data.
Velocity is a vector; acceleration can change its direction or its magnitude.
Constant-acceleration formulas are solutions under assumptions, not universal motion identities.

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