Particle kinematics in rectangular coordinates

Before a single force enters the picture, mechanics asks the purely geometric question: how does a moving point's position change with time? Particle kinematics in rectangular coordinates answers this by resolving motion along fixed, mutually perpendicular axes $x$ , $y$ , $z$ and tracking position, velocity, and acceleration as vectors built from ordinary one-dimensional calculus. It opens the Kinematics and Kinetics of Particles unit of ME10001 Mechanics, carrying forward the vector-algebra tools from Statics but shifting the question from "what forces balance?" to "how does a point move?" Rectilinear (straight-line) motion is the one-axis special case; curvilinear motion — a ball's flight, a car rounding a bend — is the general case, and rectangular coordinates are the natural choice whenever the axes of interest are straight and independent, as in projectile motion.

The core insight

Because the unit vectors $\hat i$ , $\hat j$ , $\hat k$ never change direction, differentiating the position vector $\vec r(t) = x(t)\hat i + y(t)\hat j + z(t)\hat k$ splits motion into three completely independent one-dimensional problems joined only by the shared clock $t$ .

The core idea

Rectilinear motion. Along a single axis $s$ , position, velocity, and acceleration chain together through calculus:

v = \frac{ds}{dt}, \qquad a = \frac{dv}{dt} = v\,\frac{dv}{ds}.

where: $s$ is position along the axis, $v$ is velocity, $a$ is acceleration, and $t$ is time. The second form, $a = v\,dv/ds$ , is useful whenever acceleration is known as a function of position rather than time.

Constant acceleration. Integrating the equations above under a constant $a$ gives the three formulas every projectile and free-fall problem reduces to:

v = v_0 + at, \qquad s = s_0 + v_0 t + \tfrac12 a t^2, \qquad v^2 = v_0^2 + 2a(s - s_0).

where: $v_0$ , $s_0$ are the velocity and position at $t = 0$ . These hold only while $a$ stays constant — the commonest source of wrong answers is applying them when $a$ varies with time or position.

Curvilinear motion. In three dimensions the position vector $\vec r(t) = x(t)\hat i + y(t)\hat j + z(t)\hat k$ differentiates component-by-component because $\hat i, \hat j, \hat k$ are fixed:

\vec v = \dot x\,\hat i + \dot y\,\hat j + \dot z\,\hat k, \qquad \vec a = \ddot x\,\hat i + \ddot y\,\hat j + \ddot z\,\hat k.

where: dots denote time derivatives ( $\dot x = dx/dt$ , $\ddot x = d^2x/dt^2$ ). The speed is the vector's magnitude, $v = \sqrt{\dot x^2 + \dot y^2 + \dot z^2}$ , and likewise for the magnitude of acceleration. This independence is the entire payoff of rectangular coordinates: a two-dimensional problem becomes two separate one-dimensional problems, solved on their own and recombined only through $t$ .

Derivation

Derivation AConstant-acceleration kinematic equations from calculus

The goal is to derive the three constant-acceleration formulas ( $v = v_0 + at$ , $s = s_0 + v_0 t + \tfrac12 a t^2$ , $v^2 = v_0^2 + 2a(s-s_0)$ ) directly from the defining relations $v = ds/dt$ and $a = dv/dt$ , so that every projectile, free-fall, or straight-line problem in this unit can be traced back to a single act of integration rather than memorized as disconnected rules.

Step 1. Start from the definition of acceleration, $a = dv/dt$ , and treat $a$ as constant. Separate variables and integrate both sides from $t=0$ (where $v = v_0$ ) to a general time $t$ : $\int_{v_0}^{v} dv = \int_0^t a\,dt$ . Since $a$ does not depend on $t$ , the right side is simply $a t$ , giving $v - v_0 = at$ , or $v = v_0 + at$ .

Step 2. Now use the definition of velocity, $v = ds/dt$ , substituting the result of Step 1: $ds/dt = v_0 + at$ . Integrate both sides from $t = 0$ (where $s = s_0$ ) to a general time $t$ : $\int_{s_0}^{s} ds = \int_0^t (v_0 + at)\,dt$ . The right side integrates term-by-term to $v_0 t + \tfrac12 a t^2$ , so $s - s_0 = v_0 t + \tfrac12 a t^2$ , or $s = s_0 + v_0 t + \tfrac12 a t^2$ .

Step 3. To eliminate $t$ and connect velocity directly to position, use $a = v\,dv/ds$ (the chain-rule form of acceleration). Separate variables: $v\,dv = a\,ds$ . Integrate from $s_0$ (where $v = v_0$ ) to $s$ : $\int_{v_0}^{v} v\,dv = \int_{s_0}^{s} a\,ds$ . The left side gives $\tfrac12(v^2 - v_0^2)$ and the right side gives $a(s - s_0)$ , so $v^2 - v_0^2 = 2a(s-s_0)$ , or $v^2 = v_0^2 + 2a(s-s_0)$ .

Step 4. All three results assumed only that $a$ is constant — no other physics was used. This is why the formulas fail the instant $a$ varies with time or position; in that case one must return to Step 1 and integrate $a(t)$ or $a(s)$ directly, exactly as with the cubic $s(t) = 2t^3 - 9t^2 + 12t$ from Pass 1, where $a(t) = 12t - 18$ is not constant and the shortcut formulas do not apply.

Worked example

A ball is launched from ground level with initial speed $20\ \text{m/s}$ at $30^\circ$ above the horizontal; take $g = 9.8\ \text{m/s}^2$ and neglect air resistance. Find the maximum height, the range, and the velocity (magnitude and direction) at the instant it lands.

Step 1. Resolve the launch velocity into rectangular components: $v_{0x} = 20\cos30^\circ = 17.32\ \text{m/s}$ , $v_{0y} = 20\sin30^\circ = 10.0\ \text{m/s}$ . Horizontal motion has $a_x = 0$ , so $x(t) = v_{0x}t$ ; vertical motion has $a_y = -9.8\ \text{m/s}^2$ , so $y(t) = v_{0y}t - \tfrac12 g t^2$ and $v_y(t) = v_{0y} - gt$ .

Step 2. Maximum height occurs when $v_y = 0$ : $t_{top} = v_{0y}/g = 10.0/9.8 = 1.020\ \text{s}$ , giving $y_{max} = v_{0y}t_{top} - \tfrac12 g t_{top}^2 = 10.20 - 5.10 = 5.10\ \text{m}$ .

Step 3. By the symmetry of projectile motion, total flight time is $t_f = 2t_{top} = 2.041\ \text{s}$ , so the range is $R = v_{0x}\,t_f = 17.32 \times 2.041 = 35.35\ \text{m}$ .

Step 4. At landing, $v_x = v_{0x} = 17.32\ \text{m/s}$ throughout (unchanged, since $a_x = 0$ ), while $v_y = v_{0y} - g t_f = 10.0 - 9.8(2.041) = -10.0\ \text{m/s}$ — exactly the negative of the launch value, as symmetry requires. The landing speed is $v = \sqrt{v_x^2 + v_y^2} = \sqrt{17.32^2 + 10.0^2} = \sqrt{300 + 100} = 20.0\ \text{m/s}$ , equal to the launch speed, at angle $\theta = \tan^{-1}(10.0/17.32) = 30^\circ$ below the horizontal. Maximum height $5.10\ \text{m}$ , range $35.35\ \text{m}$ , landing velocity $20.0\ \text{m/s}$ at $30^\circ$ below the horizontal.

Competitive stretch

Why must the landing speed equal the launch speed here? Because $a_x = 0$ leaves $v_x$ untouched, and $y$ returns to its starting value ( $y=0$ ) at landing, so $v^2 = v_{0y}^2 + 2a_y(y - y_0) = v_{0y}^2$ forces $|v_y|_{\text{land}} = |v_{0y}|$ by the third constant-acceleration formula from the derivation above — the launch and landing heights being equal is what makes the speeds match, and this fails immediately if the ball is launched from or lands at a different height.

Common traps

Treating $x$ - and $y$ -motion as coupled: in rectangular coordinates they are independent; the only quantity they share is time $t$ .
Dropping or flipping the sign of $g$ — if up is positive, $a_y = -9.8\ \text{m/s}^2$ for the entire flight, not just on the way down.
Confusing average velocity ( $\Delta s/\Delta t$ ) with instantaneous velocity ( $ds/dt$ ); these agree only when velocity is constant.
Applying $v = v_0 + at$ , $s = s_0 + v_0t + \tfrac12at^2$ , or $v^2 = v_0^2 + 2a(s-s_0)$ when acceleration is not constant — as in $a(t) = 12t - 18$ , where only direct integration is valid.

Ask the tutor: Want to try a variant where the projectile launches from a height, or work through integrating a non-constant $a(t)$ step by step?

Why can rectangular components be treated as totally independent, when the motion itself looks like one curved path?

The path looks curved because we usually plot $y$ against $x$ , but the equations governing $x(t)$ and $y(t)$ never reference each other — each obeys its own one-dimensional kinematics with its own acceleration. The curve you see is just the geometric trace of two unrelated clocks ticking together. This only holds because $\hat i$ and $\hat j$ are fixed in direction; in coordinate systems where the unit vectors rotate (like polar coordinates), extra terms couple the components, and this independence is lost.

How do I know whether to use the constant-acceleration formulas or go back to direct integration?

Check whether $a$ is actually constant. If a problem gives $a(t)$ or $a(s)$ as a function that is not a fixed number — like $a(t) = 12t - 18$ from the cubic-position example — the three shortcut formulas are invalid, and you must integrate $a = dv/dt$ or $a = v\,dv/ds$ directly, exactly as done in the derivation above, using the given function instead of a constant.

At the top of a projectile's flight, is the acceleration zero along with the velocity?

No — this is one of the most common mix-ups. At the peak, $v_y = 0$ because the particle is momentarily at rest vertically, but $a_y = -g$ the entire time, including at the peak. Gravity never switches off; it is precisely because $a_y$ stays nonzero that the particle immediately starts moving downward again rather than hovering.