Two dimensional motion with vectors

Motion in a plane has two independent components. A particle can move east while accelerating north, or have zero vertical velocity while still moving horizontally. Vector notation keeps these simultaneous changes together.

With fixed perpendicular unit vectors $\mathbf i$ and $\mathbf j$, usually pointing right and upwards, write the position as

$$\mathbf r=x\mathbf i+y\mathbf j.$$

The functions $x(t)$ and $y(t)$ describe the same particle at the same time $t$. Differentiate each component to obtain velocity and acceleration:

$$\boxed{\mathbf v=\frac{\mathrm d\mathbf r}{\mathrm dt} =\frac{\mathrm dx}{\mathrm dt}\mathbf i+\frac{\mathrm dy}{\mathrm dt}\mathbf j}$$

and

$$\boxed{\mathbf a=\frac{\mathrm d\mathbf v}{\mathrm dt} =\frac{\mathrm d^2\mathbf r}{\mathrm dt^2}}.$$

Prerequisites

You should be able to:

• distinguish displacement, velocity, speed and acceleration using kinematics language;
• use the constant acceleration equations;
• add vectors and find a vector’s magnitude and direction;
• differentiate and integrate polynomials using calculus in kinematics;
• solve simultaneous equations and simple quadratics.

Position and displacement vectors

Relative to an origin $O$, the position vector of a particle $P$ is

$$\overrightarrow{OP}=\mathbf r=x\mathbf i+y\mathbf j.$$

If the particle moves from position $\mathbf r_1$ to position $\mathbf r_2$, its displacement is

$$\boxed{\Delta\mathbf r=\mathbf r_2-\mathbf r_1}.$$

Subtract initial position from final position. The displacement is not generally the final position vector, and its magnitude is not necessarily the distance travelled.

Worked example 1: position, displacement and distance

A particle moves from $A$, with position vector $(2\mathbf i-3\mathbf j)\ \mathrm m$, to $B$, with position vector $(8\mathbf i+5\mathbf j)\ \mathrm m$, along the straight line segment $AB$. Find its displacement and the distance travelled.

The displacement is

$$\begin{aligned} \overrightarrow{AB} &=(8\mathbf i+5\mathbf j)-(2\mathbf i-3\mathbf j)\\ &=6\mathbf i+8\mathbf j\ \mathrm m. \end{aligned}$$

Because the path is the straight segment from $A$ to $B$, the distance travelled equals the magnitude of this displacement:

$$|\overrightarrow{AB}|=\sqrt{6^2+8^2}=10\ \mathrm m.$$

Thus the displacement is $\boxed{(6\mathbf i+8\mathbf j)\ \mathrm m}$ and the distance is $\boxed{10\ \mathrm m}$.

If the route from $A$ to $B$ had curved or doubled back, the displacement would be unchanged but the distance would exceed $10$ m.

Self-check 1

A particle moves from position $(-4\mathbf i+7\mathbf j)\ \mathrm m$ to $(5\mathbf i-5\mathbf j)\ \mathrm m$. Find its displacement and the magnitude of its displacement.

Answer

$$\Delta\mathbf r=(5-(-4))\mathbf i+(-5-7)\mathbf j =9\mathbf i-12\mathbf j\ \mathrm m.$$

Therefore

$$|\Delta\mathbf r|=\sqrt{9^2+(-12)^2}=\boxed{15\ \mathrm m}.$$

From position to velocity and acceleration

Suppose

$$\mathbf r(t)=x(t)\mathbf i+y(t)\mathbf j.$$

Since $\mathbf i$ and $\mathbf j$ are fixed, differentiate the scalar components separately:

$$\mathbf v(t)=x'(t)\mathbf i+y'(t)\mathbf j,$$ $$\mathbf a(t)=x''(t)\mathbf i+y''(t)\mathbf j.$$

The component $x'(t)$ is the rate at which the horizontal coordinate changes. It is not the speed unless $y'(t)=0$.

The speed is the magnitude of velocity:

$$\boxed{|\mathbf v|=\sqrt{v_x^2+v_y^2}}.$$

If velocity is nonzero, it points along the tangent to the path in the direction of motion.

Worked example 2: differentiate a position vector

A particle has position vector

$$\mathbf r(t)=(t^2+2t)\mathbf i+(12t-t^3)\mathbf j\ \mathrm m, \qquad t\geq0.$$

Find its velocity, acceleration and speed at $t=2$.

Differentiate componentwise:

$$\mathbf v(t)=(2t+2)\mathbf i+(12-3t^2)\mathbf j\ \mathrm{m\,s^{-1}}.$$

Differentiate again:

$$\mathbf a(t)=2\mathbf i-6t\mathbf j\ \mathrm{m\,s^{-2}}.$$

At $t=2$,

$$\mathbf v(2)=6\mathbf i+0\mathbf j=6\mathbf i\ \mathrm{m\,s^{-1}},$$ $$\mathbf a(2)=2\mathbf i-12\mathbf j\ \mathrm{m\,s^{-2}}.$$

Therefore the speed is

$$|\mathbf v(2)|=\sqrt{6^2+0^2}=\boxed{6\ \mathrm{m\,s^{-1}}}.$$

At this instant the particle moves horizontally, but its acceleration is not horizontal. In particular, zero vertical velocity does not imply zero vertical acceleration.

Velocity and acceleration need not point together

Velocity describes the particle’s present motion. Acceleration describes how velocity is changing. A particle may accelerate perpendicular to its velocity, as in circular motion, or opposite to one velocity component but not the other.

Self-check 2

The position of a particle is

$$\mathbf r(t)=(3t^2-1)\mathbf i+(2t^3-5t)\mathbf j\ \mathrm m.$$

Find $\mathbf v$, $\mathbf a$ and the speed at $t=1$.

Answer

$$\mathbf v=6t\mathbf i+(6t^2-5)\mathbf j, \qquad \mathbf a=6\mathbf i+12t\mathbf j.$$

At $t=1$,

$$\mathbf v(1)=6\mathbf i+\mathbf j, \qquad \mathbf a(1)=6\mathbf i+12\mathbf j.$$

Hence the speed is

$$\boxed{\sqrt{37}\ \mathrm{m\,s^{-1}}}.$$

Magnitude and direction of motion

For

$$\mathbf v=v_x\mathbf i+v_y\mathbf j,$$

the speed is $\sqrt{v_x^2+v_y^2}$. To find direction, first use the signs of $v_x$ and $v_y$ to identify the quadrant. Then use a positive reference angle:

$$\tan\alpha=\frac{|v_y|}{|v_x|}.$$

Do not trust $\tan^{-1}(v_y/v_x)$ alone to describe the quadrant. If bearings are requested, remember that bearings are measured clockwise from north and written with three figures.

Worked example 3: speed and bearing

A particle has velocity

$$\mathbf v=-12\mathbf i+5\mathbf j\ \mathrm{m\,s^{-1}},$$

where $\mathbf i$ points east and $\mathbf j$ points north. Find its speed and bearing.

Its speed is

$$|\mathbf v|=\sqrt{(-12)^2+5^2}=13\ \mathrm{m\,s^{-1}}.$$

The particle moves west and north, so its direction is north of west. If $\alpha$ is the angle north of west,

$$\tan\alpha=\frac{5}{12},$$

giving $\alpha\approx22.6^\circ$.

West has bearing $270^\circ$. Moving $22.6^\circ$ towards north gives

$$270^\circ+22.6^\circ=292.6^\circ.$$

The speed is $\boxed{13\ \mathrm{m\,s^{-1}}}$ on a bearing of $\boxed{293^\circ}$ to the nearest degree.

Constant acceleration in a plane

The one dimensional SUVAT equations extend directly to vectors when the acceleration vector is constant:

$$\boxed{\mathbf v=\mathbf u+\mathbf a t}$$ $$\boxed{\mathbf r=\mathbf r_0+\mathbf u t+\frac12\mathbf a t^2}.$$

Equivalently, resolve into perpendicular directions:

$$v_x=u_x+a_xt, \qquad v_y=u_y+a_yt,$$ $$x=x_0+u_xt+\frac12a_xt^2, \qquad y=y_0+u_yt+\frac12a_yt^2.$$

The same $t$ appears in both directions because these components describe one motion. The axes are mathematically independent, but the clock is shared.

What constant acceleration means

The whole vector $\mathbf a$ is constant only if both $a_x$ and $a_y$ are constant. Velocity may still change in magnitude, direction, or both.

Worked example 4: constant vector acceleration

At $t=0$, a particle is at

$$\mathbf r_0=(3\mathbf i-2\mathbf j)\ \mathrm m$$

with velocity

$$\mathbf u=(4\mathbf i+7\mathbf j)\ \mathrm{m\,s^{-1}}.$$

It has constant acceleration

$$\mathbf a=(2\mathbf i-3\mathbf j)\ \mathrm{m\,s^{-2}}.$$

Find its position and velocity after $4$ seconds, and its speed then.

For velocity,

$$\begin{aligned} \mathbf v &=\mathbf u+\mathbf a t\\ &=(4\mathbf i+7\mathbf j)+4(2\mathbf i-3\mathbf j)\\ &=12\mathbf i-5\mathbf j\ \mathrm{m\,s^{-1}}. \end{aligned}$$

For position,

$$\begin{aligned} \mathbf r &=\mathbf r_0+\mathbf u t+\frac12\mathbf a t^2\\ &=(3\mathbf i-2\mathbf j)+4(4\mathbf i+7\mathbf j) +\frac12(2\mathbf i-3\mathbf j)(4^2)\\ &=35\mathbf i+2\mathbf j\ \mathrm m. \end{aligned}$$

The speed after $4$ seconds is

$$|\mathbf v|=\sqrt{12^2+(-5)^2}=\boxed{13\ \mathrm{m\,s^{-1}}}.$$

Thus $\boxed{\mathbf r=35\mathbf i+2\mathbf j\ \mathrm m}$ and $\boxed{\mathbf v=12\mathbf i-5\mathbf j\ \mathrm{m\,s^{-1}}}$.

Self-check 3

A particle starts at the origin with velocity $(6\mathbf i-2\mathbf j)\ \mathrm{m\,s^{-1}}$ and constant acceleration $(-\mathbf i+4\mathbf j)\ \mathrm{m\,s^{-2}}$. Find its position, velocity and speed after $3$ seconds.

Answer

$$\begin{aligned} \mathbf r(3) &=3(6\mathbf i-2\mathbf j)+\frac12(3^2)(-\mathbf i+4\mathbf j)\\ &=\frac{27}{2}\mathbf i+12\mathbf j\ \mathrm m. \end{aligned}$$

Also,

$$\mathbf v(3)=(6\mathbf i-2\mathbf j)+3(-\mathbf i+4\mathbf j) =3\mathbf i+10\mathbf j\ \mathrm{m\,s^{-1}}.$$

Therefore the speed is

$$\boxed{\sqrt{109}\ \mathrm{m\,s^{-1}}}.$$

Finding when a particle has a given direction

A direction condition is a relationship between velocity components. For example:

• motion parallel to $\mathbf i$ means $v_y=0$;
• motion parallel to $\mathbf j$ means $v_x=0$;
• motion at $45^\circ$ above positive $\mathbf i$ means $v_y=v_x>0$;
• motion parallel to $p\mathbf i+q\mathbf j$ means $\mathbf v=\lambda(p\mathbf i+q\mathbf j)$ for some scalar $\lambda$.

The scalar $\lambda$ matters. If $\lambda<0$, the velocity is parallel to the same line but points in the opposite sense.

Worked example 5: direction as a component condition

A particle has velocity

$$\mathbf v(t)=(10-2t)\mathbf i+(1+3t)\mathbf j\ \mathrm{m\,s^{-1}}.$$

Find when it moves parallel to $2\mathbf i+5\mathbf j$, and state whether it moves in the same or opposite sense.

Parallel vectors have proportional components, so

$$\frac{10-2t}{2}=\frac{1+3t}{5}.$$

Cross multiply:

$$5(10-2t)=2(1+3t).$$

Hence

$$50-10t=2+6t, \qquad t=3.$$

At $t=3$,

$$\mathbf v=4\mathbf i+10\mathbf j=2(2\mathbf i+5\mathbf j).$$

The multiplier is positive, so the particle moves in the same sense. This occurs at $\boxed{t=3\ \mathrm s}$.

Recovering position from velocity or acceleration

Integrate each component separately. Every integration introduces constants, determined from initial conditions.

If

$$\mathbf v(t)=v_x(t)\mathbf i+v_y(t)\mathbf j,$$

then

$$\mathbf r(t)=\left(\int v_x(t)\,\mathrm dt\right)\mathbf i +\left(\int v_y(t)\,\mathrm dt\right)\mathbf j.$$

It is often cleanest to use a constant vector $\mathbf c=C_1\mathbf i+C_2\mathbf j$.

Worked example 6: integrate and use an initial position

A particle has velocity

$$\mathbf v(t)=3t^2\mathbf i+(4-2t)\mathbf j\ \mathrm{m\,s^{-1}}.$$

At $t=0$, its position is $2\mathbf i-3\mathbf j$ m. Find its position at time $t$, then find when it next crosses the horizontal line $y=0$.

Integrate:

$$\mathbf r(t)=(t^3+C_1)\mathbf i+(4t-t^2+C_2)\mathbf j.$$

Use $\mathbf r(0)=2\mathbf i-3\mathbf j$:

$$C_1=2, \qquad C_2=-3.$$

Therefore

$$\boxed{\mathbf r(t)=(t^3+2)\mathbf i+(4t-t^2-3)\mathbf j\ \mathrm m}.$$

On $y=0$,

$$4t-t^2-3=0,$$

so

$$t^2-4t+3=0 \quad\Longrightarrow\quad (t-1)(t-3)=0.$$

The particle crosses the line at $t=1$ and $t=3$. The first crossing after $t=0$ is $t=1$, and it crosses again at $t=3$.

Keep both constants

Two component integrations normally require two scalar constants. A given position vector supplies both conditions. Do not silently set either constant to zero.

The path of a particle

The vector equation

$$\mathbf r(t)=x(t)\mathbf i+y(t)\mathbf j$$

is a parametric description of the path. To find a Cartesian equation, eliminate $t$ between $x=x(t)$ and $y=y(t)$.

Worked example 7: eliminate time

A particle moves according to

$$\mathbf r(t)=2t\mathbf i+(t^2+1)\mathbf j.$$

Find the equation of its path and the velocity when it passes through $(4,5)$.

From the horizontal component,

$$x=2t \quad\Longrightarrow\quad t=\frac{x}{2}.$$

Substitute into $y=t^2+1$:

$$\boxed{y=\frac{x^2}{4}+1}.$$

Differentiate the position:

$$\mathbf v=2\mathbf i+2t\mathbf j.$$

At $(4,5)$, $x=2t$ gives $t=2$. Therefore

$$\boxed{\mathbf v=2\mathbf i+4\mathbf j}.$$

The gradient of the path there is

$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt} =\frac{4}{2}=2,$$

which agrees with the direction ratio $v_y/v_x=4/2$.

Self-check 4

A particle has position

$$\mathbf r(t)=(t+1)\mathbf i+(t^2-2t)\mathbf j.$$

Find a Cartesian equation of its path and its velocity when $x=4$.

Answer

Since $x=t+1$, $t=x-1$. Therefore

$$y=(x-1)^2-2(x-1)=\boxed{x^2-4x+3}.$$

Also,

$$\mathbf v=\mathbf i+(2t-2)\mathbf j.$$

When $x=4$, $t=3$, so

$$\boxed{\mathbf v=\mathbf i+4\mathbf j}.$$

Meeting and crossing

Two particles meet only if they have the same position at the same time. Solve

$$\mathbf r_A(t)=\mathbf r_B(t),$$

which means solving both component equations for one common value of $t$.

Their paths can cross geometrically without the particles colliding. A shared point reached at different times is not a meeting.

Worked example 8: do two particles meet?

Particles $A$ and $B$ have positions

$$\mathbf r_A=(1+2t)\mathbf i+(7-t)\mathbf j,$$ $$\mathbf r_B=(10-t)\mathbf i+(1+t)\mathbf j,$$

where $t$ is measured in seconds. Determine whether they meet.

Equal $\mathbf i$ components require

$$1+2t=10-t,$$

so $3t=9$ and $t=3$.

Check the $\mathbf j$ components at this same time:

$$7-3=4, \qquad 1+3=4.$$

Both components agree. Therefore the particles meet after $\boxed{3\ \mathrm s}$ at

$$\boxed{7\mathbf i+4\mathbf j}.$$

If the second component equation had produced a different time, the particles would not meet.

Common misconceptions

Treating components as separate journeys

The $x$ and $y$ equations use one shared time. They are two views of the same motion, not two particles or two clocks.

Confusing speed with a velocity component

For $\mathbf v=3\mathbf i+4\mathbf j$, neither component is the speed. The speed is $|\mathbf v|=5$.

Assuming acceleration follows motion

A particle can move north while accelerating east. Acceleration follows the change in velocity, not necessarily the current velocity.

Using position instead of displacement

Position depends on the chosen origin. Displacement over an interval is final position minus initial position.

Losing direction when using squared equations

An equation involving $v_x^2$ or $v_y^2$ gives a magnitude. Use the context or another equation to determine the sign of the component.

Equating paths with collisions

Intersecting paths show only that both particles can occupy the same point. A collision also requires equal times.

Mixed self-check

A particle has acceleration

$$\mathbf a=2\mathbf i-4\mathbf j\ \mathrm{m\,s^{-2}}.$$

At $t=0$, its velocity is $-2\mathbf i+8\mathbf j\ \mathrm{m\,s^{-1}}$ and its position is $3\mathbf i+\mathbf j$ m.

1. Find $\mathbf v(t)$ and $\mathbf r(t)$.
2. Find when the particle moves horizontally.
3. Find its position and speed then.

Answer

Integrating acceleration, or using constant acceleration equations,

$$\mathbf v(t)=(-2+2t)\mathbf i+(8-4t)\mathbf j.$$

Then

$$\mathbf r(t)=(3-2t+t^2)\mathbf i+(1+8t-2t^2)\mathbf j.$$

Horizontal motion requires the vertical velocity component to be zero:

$$8-4t=0 \quad\Longrightarrow\quad t=2\ \mathrm s.$$

At this time,

$$\mathbf r(2)=3\mathbf i+9\mathbf j\ \mathrm m,$$

and

$$\mathbf v(2)=2\mathbf i\ \mathrm{m\,s^{-1}}.$$

Therefore the speed is $\boxed{2\ \mathrm{m\,s^{-1}}}$ and the position is $\boxed{3\mathbf i+9\mathbf j\ \mathrm m}$.

Exam checklist

Before accepting a solution, check that you have:

1. defined the directions of $\mathbf i$ and $\mathbf j$ if they are not given;
2. used one common time in both components;
3. subtracted positions in the correct order for displacement;
4. differentiated or integrated every component;
5. used $|\mathbf v|$, not one component, for speed;
6. checked the quadrant before stating a direction;
7. included units and interpreted negative components;
8. checked both coordinates when deciding whether particles meet.

Next steps

Apply this component method to projectile motion, where horizontal acceleration is zero and vertical acceleration is $-g$. Then connect forces to acceleration in dynamics in a plane. For motion given by more general vector functions, consolidate calculus in kinematics.