Motion in two dimensions with vectorsAQA A-Level Maths: Revision notes

In two dimensions, position, velocity and acceleration are vectors written in terms of perpendicular unit vectors $\mathbf i$ and $\mathbf j$:

• position vector $\mathbf r=x\mathbf i+y\mathbf j$, measured from a fixed origin $O$;
• velocity $\mathbf v=v_x\mathbf i+v_y\mathbf j$;
• acceleration $\mathbf a=a_x\mathbf i+a_y\mathbf j$. Speed is the magnitude of the velocity, $\sqrt{v_x^2+v_y^2}$, and is a scalar. The direction of motion is given by the angle $\theta$ with $\tan\theta=\frac{v_y}{v_x}$. The displacement between two times is the change in position vector, and the distance from $O$ is $\lvert\mathbf r\rvert=\sqrt{x^2+y^2}$. Treat the $\mathbf i$ and $\mathbf j$ components separately.

The constant acceleration formulae extend to vectors, with $\mathbf s$ the displacement: $\mathbf v=\mathbf u+\mathbf at,\qquad\mathbf s=\mathbf ut+\tfrac12\mathbf at^2,\qquad\mathbf s=\tfrac12(\mathbf u+\mathbf v)t,\qquad\mathbf s=\mathbf vt-\tfrac12\mathbf at^2.$ The position at time $t$ is $\mathbf r=\mathbf r_0+\mathbf s$, where $\mathbf r_0$ is the initial position. Each equation applies separately to the $\mathbf i$ and $\mathbf j$ components. Example: $\mathbf a=2\mathbf i-\mathbf j$, $\mathbf u=\mathbf i+3\mathbf j$, $\mathbf r_0=3\mathbf i+4\mathbf j$. At $t=3$: $\mathbf v=(\mathbf i+3\mathbf j)+3(2\mathbf i-\mathbf j)=7\mathbf i$, and $\mathbf r=(3\mathbf i+4\mathbf j)+3(\mathbf i+3\mathbf j)+4.5(2\mathbf i-\mathbf j)=15\mathbf i+8.5\mathbf j$.

When the motion is described by functions of time, use calculus. Differentiate each component: $\mathbf v=\frac{d\mathbf r}{dt},\qquad\mathbf a=\frac{d\mathbf v}{dt}=\frac{d^2\mathbf r}{dt^2}.$ Example: $\mathbf r=(t^3-3t)\mathbf i+(4t-t^2)\mathbf j$ gives $\mathbf v=(3t^2-3)\mathbf i+(4-2t)\mathbf j$ and $\mathbf a=6t\,\mathbf i-2\mathbf j$. At $t=3$, $\mathbf v=24\mathbf i-2\mathbf j$ and at $t=2$, $\mathbf a=12\mathbf i-2\mathbf j$.

Reverse the process by integrating each component: $\mathbf v=\int\mathbf a\,dt,\qquad\mathbf r=\int\mathbf v\,dt.$ Each integration gives a vector constant of integration $\mathbf c$, found from given conditions such as the initial velocity or initial position. Example: $\mathbf a=4t\,\mathbf i-2\mathbf j$ with $\mathbf v=2\mathbf i+5\mathbf j$ at $t=0$. Then $\mathbf v=2t^2\mathbf i-2t\mathbf j+\mathbf c$, and $\mathbf c=2\mathbf i+5\mathbf j$, so $\mathbf v=(2t^2+2)\mathbf i+(5-2t)\mathbf j$. If also $\mathbf r=\mathbf i+3\mathbf j$ at $t=0$, then $\mathbf r=\left(\frac23t^3+2t+1\right)\mathbf i+(5t-t^2+3)\mathbf j$.

• At rest: the velocity is zero, which needs both components to be zero at the same time. Solve one component, then test the other.
• Moving parallel to $\mathbf i$: the $\mathbf j$-component of velocity is zero (and the $\mathbf i$-component is not). Similarly for $\mathbf j$.
• Speed: the magnitude of $\mathbf v$.
• Distance from $O$: the magnitude of $\mathbf r$. Example: $\mathbf v=(3t^2-12t+9)\mathbf i+(4t-8)\mathbf j$. The $\mathbf j$-component is zero only at $t=2$, where the $\mathbf i$-component is $-3$, so the particle is never at rest. At $t=2$ it moves parallel to $\mathbf i$, with speed $3$ m s$^{-1}$.