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Mechanics · Kinematics

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Motion in two dimensions with vectors

Why the a → v → r calculus chain works component-by-component in 2D, that speed is the magnitude of the velocity vector, and how "parallel to" and "perpendicular to" a direction become simple component conditions.

Mechanics · Kinematics

Motion in two dimensions with vectors

Why the a → v → r calculus chain works component-by-component in 2D, that speed is the magnitude of the velocity vector, and how "parallel to" and "perpendicular to" a direction become simple component conditions.

Why it works

In two dimensions a particle's position, velocity and acceleration are all vectors, written in i\mathbf{i}, j\mathbf{j} components. The good news: nothing about the calculus changes. If the position is r=x(t) i+y(t) j,\mathbf{r} = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}, then differentiating gives the velocity and acceleration one component at a time: v=drdt=x˙ i+y˙ j,a=dvdt=x¨ i+y¨ j.\mathbf{v} = \frac{d\mathbf{r}}{dt} = \dot{x}\,\mathbf{i} + \dot{y}\,\mathbf{j}, \qquad \mathbf{a} = \frac{d\mathbf{v}}{dt} = \ddot{x}\,\mathbf{i} + \ddot{y}\,\mathbf{j}. The i\mathbf{i}-component carries its own one-dimensional motion and the j\mathbf{j}-component carries its own; they don't interact. Integration goes back up the chain the same way, with a vector constant fixed by the initial conditions (the constant for v\mathbf{v} is the initial velocity vector, for r\mathbf{r} the initial position). So 2D kinematics is just the straight-line calculus of [[kinematics.integrating-motion]] done twice in parallel.

Speed is a scalar — the magnitude of the velocity vector. This catches people out: "find the speed" never wants the vector v\mathbf{v}, it wants its size, speed=∣v∣=vx2+vy2.\text{speed} = |\mathbf{v}| = \sqrt{v_x^2 + v_y^2}. A subtlety worth its own warning: the acceleration is not the rate of change of the speed. Differentiate the components of v\mathbf{v} to get a\mathbf{a}; do not differentiate ∣v∣|\mathbf{v}|.

Direction conditions become component conditions. Two phrases come up constantly, and both reduce to looking at components:
  • "moving parallel to a direction d\mathbf{d}" means v\mathbf{v} is a scalar
multiple of d\mathbf{d}, so their components are in the same ratio. For d=i+j\mathbf{d} = \mathbf{i} + \mathbf{j} that means the i\mathbf{i}- and j\mathbf{j}-components of v\mathbf{v} are equal; for d=i\mathbf{d} = \mathbf{i} it means the j\mathbf{j}-component is zero (the motion is purely horizontal).
  • "perpendicular to i\mathbf{i}" means the velocity (or acceleration) has **no
i\mathbf{i}-component, i.e. its i\mathbf{i}-component is zero and it points purely along j\mathbf{j}. (Likewise "perpendicular to j\mathbf{j}" sets the j\mathbf{j}-component to zero.)

So a question like "find tt when the acceleration is perpendicular to i\mathbf{i}" is solved by setting the i\mathbf{i}-component of a\mathbf{a} to zero and solving for tt — no new machinery.

Constant acceleration in 2D is the special case where a\mathbf{a} doesn't depend on tt, and then the suvat equations hold as vector equations**: v=u+at,r=ut+12at2.\mathbf{v} = \mathbf{u} + \mathbf{a}t, \qquad \mathbf{r} = \mathbf{u}t + \tfrac12\mathbf{a}t^2. The same warning as in one dimension applies, twice over: the moment a\mathbf{a} varies with tt, these break and you must integrate component-by-component instead.