Kinematics in One and Three Dimensions
~6 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)
You should be comfortable with basic vector notation, trigonometric identities, and the concept of a derivative as an instantaneous rate of change.
Why this matters
Kinematics provides the precise language to describe motion before we ask what causes it.
Before we can apply Newton's laws to predict how forces change a particle's trajectory, we must rigorously define position, velocity, and acceleration. This lecture establishes the mathematical framework for one-dimensional and three-dimensional motion, culminating in the analysis of projectile motion and uniform circular motion, which serve as foundational models for more complex dynamical systems.
The Geometry of Motion
Imagine tracking the position of a particle moving through space. We define its location using a position vector \(\vec{r}(t)\) relative to an origin. The change in this position over time defines velocity, while the change in velocity over time defines acceleration. In one dimension, these are simple scalar quantities with signs indicating direction. In three dimensions, they are vectors, requiring us to track components along orthogonal axes. A useful insight is that, when the acceleration is constant, each Cartesian component can be treated on its own: the horizontal motion of a projectile is unaffected by its vertical motion, which turns a curved trajectory into two simple straight-line problems.
Constant Acceleration in One Dimension
When acceleration is constant, the equations of motion become particularly simple. Starting from the definition of acceleration \(a = \frac{dv}{dt}\), integration yields the velocity as a linear function of time: \(v(t) = v_0 + at\). Integrating again gives the position as a quadratic function: \(x(t) = x_0 + v_0 t + \frac{1}{2}at^2\). These equations allow us to solve for any unknown variable if three others are known. They form the basis for analyzing free fall, where \(a = -g\), and other uniformly accelerated systems.
Projectile Motion as Composite Motion
A projectile launched with initial speed \(v_0\) at an angle \(\theta\) above the horizontal experiences only gravitational acceleration \(\vec{a} = -g\,\hat{y}\). By decomposing the initial velocity into horizontal \(v_{0x} = v_0 \cos\theta\) and vertical \(v_{0y} = v_0 \sin\theta\) components, we treat the motion as two independent one-dimensional problems. The horizontal motion has zero acceleration, resulting in constant velocity. The vertical motion has constant downward acceleration. Eliminating time \(t\) from the two equations yields the parabolic trajectory:
Key derived quantities include the time of flight \(T = \frac{2v_0\sin\theta}{g}\), maximum height \(H = \frac{v_0^2\sin^2\theta}{2g}\), and range \(R = \frac{v_0^2\sin 2\theta}{g}\). Note that the range is maximized when \(\theta = 45^\circ\), as \(\sin 2\theta\) reaches its peak value of 1.
Uniform Circular Motion
Consider a particle moving in a circle of radius \(R\) with constant speed \(v\). Although the speed is constant, the direction of the velocity vector changes continuously, resulting in a non-zero acceleration. This acceleration is directed radially inward toward the center of the circle and is called centripetal acceleration. Its magnitude is given by \(a_c = \frac{v^2}{R}\). In terms of angular speed \(\omega\), where \(v = \omega R\), the acceleration is \(a_c = \omega^2 R\). According to Newton's second law, a net force \(F_c = m a_c\) must act inward to maintain this circular path. The period of revolution is \(T = \frac{2\pi R}{v} = \frac{2\pi}{\omega}\).
The derivations
Projectile Motion from Constant-Acceleration Kinematics
For a particle of constant mass \(m\) moving under a constant acceleration \(\vec{a}\), the position and velocity are linear and quadratic functions of time: \(\vec{v}(t)=\vec{v}_0+\vec{a}t\) and \(\vec{r}(t)=\vec{r}_0+\vec{v}_0 t+\tfrac{1}{2}\vec{a}t^2\). When this acceleration is the uniform gravitational field \(\vec{a}=-g\,\hat{y}\) near the Earth's surface and air resistance is neglected, a projectile launched with speed \(v_0\) at angle \(\theta\) above the horizontal traces the parabola \(y(x)=x\tan\theta-\dfrac{g\,x^2}{2v_0^2\cos^2\theta}\), with time of flight \(T=\dfrac{2v_0\sin\theta}{g}\), maximum height \(H=\dfrac{v_0^2\sin^2\theta}{2g}\), and range on level ground \(R=\dfrac{v_0^2\sin 2\theta}{g}\).
Uniform Circular Motion and Centripetal Acceleration
For a particle of mass \(m\) moving in a circle of radius \(R\) with constant angular speed \(\omega\), the acceleration is directed radially inward with magnitude \(a_c = \omega^2 R = v^2/R\), where \(v = \omega R\) is the constant speed; Newton's second law then requires a net inward force \(F_c = m v^2/R = m \omega^2 R\), and the period of revolution is \(T = 2\pi/\omega = 2\pi R/v\).
A common misconception is that centripetal force is a new type of force distinct from gravity, tension, or friction. In reality, it is merely the name for the net force component directed toward the center of curvature. For example, in a vertical loop, the centripetal force is provided by the combination of gravity and normal reaction, which vary in magnitude and direction throughout the motion. Confusing the cause (specific forces) with the effect (centripetal acceleration) leads to incorrect free-body diagrams. C
Worked example
Problem. A ball is launched from ground level with an initial speed of \(20\ \mathrm{m/s}\) at an angle of \(30^\circ\) above the horizontal. Neglecting air resistance, calculate the maximum height reached and the total time of flight. Use \(g = 9.8\ \mathrm{m/s^2}\).
- Identify the initial vertical velocity component: \(v_{0y} = v_0 \sin\theta = 20 \sin 30^\circ = 20 \times 0.5 = 10\ \mathrm{m/s}\).
- Calculate the maximum height \(H\) using the kinematic relation \(v_y^2 = v_{0y}^2 - 2gH\). At the peak, \(v_y = 0\), so \(0 = 10^2 - 2(9.8)H\). Solving for \(H\): \(H = \frac{100}{19.6} \approx 5.10\ \mathrm{m}\).
- Calculate the total time of flight \(T\). The time to reach the peak is \(t_{peak} = \frac{v_{0y}}{g} = \frac{10}{9.8} \approx 1.02\ \mathrm{s}\). Since the trajectory is symmetric, the total time is \(T = 2 t_{peak} = \frac{2v_{0y}}{g} = \frac{20}{9.8} \approx 2.04\ \mathrm{s}\).
Units check. Height is in metres, time is in seconds. The formula \(H = \frac{v^2}{2g}\) yields \(\frac{\mathrm{m^2/s^2}}{\mathrm{m/s^2}} = \mathrm{m}\), which is consistent.
Three quick tests
- What is the direction of the acceleration vector for a projectile at its highest point?
Answer
Vertically downward, with magnitude \(g\). - If the launch angle for a projectile is doubled, how does the range change (assuming \(v_0\) is constant and initial angle is \(30^\circ\))?
Answer
It stays the same. The range is proportional to \(\sin 2\theta\), and \(\sin 60^\circ = \sin 120^\circ \approx 0.866\), so launch angles of \(30^\circ\) and \(60^\circ\) (complementary angles) give equal ranges on level ground. - In uniform circular motion, if the speed is doubled while the radius remains constant, what happens to the centripetal acceleration?
Answer
It quadruples, since \(a_c \propto v^2\).
Problem set for PU-101 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.
Derivations cited: D-420 · D-421 · Threads: force, energy, matter