Mk1, Mk2, Mv1, Mv2 - Position and velocity vectors

Mk1, Mk2, Mv1, Mv2 - Position and velocity vectors

Describe motion in two and three dimensions without confusing a vector with its magnitude. You will use constant-acceleration formulae, differentiate and integrate components for variable motion, infer the resultant force, and compare positions of moving particles.

Position, travel and vector magnitudes

A robot leaves the origin, travels one metre left, then one metre right. How far away is it, and how far has it travelled?

It is back at the origin, so its distance from the origin and its displacement magnitude are zero. Its distance travelled is 2 m2\,\mathrm{m}. Returning to the start does not erase the journey.

You already know

A position vector r\mathbf{r} locates a point relative to an origin. Displacement is final position minus initial position. For v=ai+bj+ck\mathbf{v}=a\mathbf{i}+b\mathbf{j}+c\mathbf{k}, speed is ∣v∣=a2+b2+c2|\mathbf{v}|=\sqrt{a^2+b^2+c^2}.

A vector tells you direction as well as magnitude. Velocity is the rate of change of position; acceleration is the rate of change of velocity. A non-zero acceleration need not increase speed: it can turn a velocity, or slow it down.

Distance from and distance travelled

The distance from a fixed point is the magnitude of the displacement from that point. Distance travelled is the length accumulated along the route. Similarly, speed is the magnitude of velocity, while magnitude of acceleration is the length of the acceleration vector.

The distinction survives in space. A particle can have a zero velocity component while still moving in the other directions. It is at rest only when every component is zero.

Key idea

Differentiate or integrate the components of a vector. Take a magnitude only when the question asks for a scalar such as speed, distance or magnitude of acceleration.

A different route checks whether you can distinguish position from travel.

Constant acceleration in components

Under a constant acceleration vector, each coordinate obeys the same one-dimensional constant-acceleration relationships.

Constant-acceleration vectors

For constant a\mathbf{a}, initial position r0\mathbf{r}_0 and initial velocity u\mathbf{u},

v=u+ta,r=r0+tu+12t2a\mathbf{v}=\mathbf{u}+t\mathbf{a},\qquad \mathbf{r}=\mathbf{r}_0+t\mathbf{u}+\frac12t^2\mathbf{a}

The displacement is r−r0\mathbf{r}-\mathbf{r}_0.

In the formula booklet.

Do not replace a vector by its magnitude in these equations. The components may change in different ways.

Worked example 1: a three-dimensional position

Question. A particle has r0=(0,1,−1) m\mathbf{r}_0=(0,1,-1)\,\mathrm{m}, u=(1,2,3) m s−1\mathbf{u}=(1,2,3)\,\mathrm{m\,s^{-1}}, and constant a=(2,−1,0) m s−2\mathbf{a}=(2,-1,0)\,\mathrm{m\,s^{-2}}. Find its position, velocity and speed at t=2 st=2\,\mathrm{s}.

Plan. Substitute t=2t=2 in the vector equations, then take the velocity magnitude.

r=(0,1,−1)+2(1,2,3)+2(2,−1,0)position formula=(6,3,5)add componentsv=(1,2,3)+2(2,−1,0)=(5,0,3)velocity formula∣v∣=25+0+9=34speed\begin{aligned} \mathbf{r}&=(0,1,-1)+2(1,2,3)+2(2,-1,0) && \text{position formula} \\ &=(6,3,5) && \text{add components} \\ \mathbf{v}&=(1,2,3)+2(2,-1,0)=(5,0,3) && \text{velocity formula} \\ |\mathbf{v}|&=\sqrt{25+0+9}=\sqrt{34} && \text{speed} \end{aligned}

Why this step? The zero second velocity component does not mean rest: the first and third components are non-zero.

Answer. Position (6,3,5) m(6,3,5)\,\mathrm{m}, velocity (5,0,3) m s−1(5,0,3)\,\mathrm{m\,s^{-1}}, speed 34 m s−1\sqrt{34}\,\mathrm{m\,s^{-1}}.

Check. With zero third acceleration, the third position is −1+3(2)=5-1+3(2)=5.

Change only the time to 3 s3\,\mathrm{s}. Keep the initial position in the position calculation.

Worked example 2: a later instant

Question. Find the same quantities at t=3 st=3\,\mathrm{s}.

Plan. Re-evaluate both vector equations using the new time.

r=(0,1,−1)+3(1,2,3)+92(2,−1,0)new time=(12,52,8)componentsv=(1,2,3)+3(2,−1,0)=(7,−1,3)velocity∣v∣=49+1+9=59magnitude\begin{aligned} \mathbf{r}&=(0,1,-1)+3(1,2,3)+\frac92(2,-1,0) && \text{new time} \\ &=(12,\tfrac52,8) && \text{components} \\ \mathbf{v}&=(1,2,3)+3(2,-1,0)=(7,-1,3) && \text{velocity} \\ |\mathbf{v}|&=\sqrt{49+1+9}=\sqrt{59} && \text{magnitude} \end{aligned}

Why this step? Position contains t2t^2; increasing time by a factor does not generally increase displacement by that factor.

Answer. (12,52,8) m(12,\frac52,8)\,\mathrm{m}, (7,−1,3) m s−1(7,-1,3)\,\mathrm{m\,s^{-1}}, and 59 m s−1\sqrt{59}\,\mathrm{m\,s^{-1}}.

Check. The second coordinate is 1+2(3)−12(32)=5/21+2(3)-\frac12(3^2)=5/2.

These vector equations require constant acceleration. For a varying acceleration, use calculus instead.

Watch out

Wrong: apply the scalar relationship v2=u2+2asv^2=u^2+2as to vector symbols as though squaring a vector gave another vector. Right: use the valid vector equations above or work with scalar components.

The next data change the initial vectors, while leaving the method intact.

Variable motion and the resultant force

Differentiation moves from position to velocity to acceleration; integration reverses that chain, with a vector constant at each stage.

The calculus chain

For differentiable vector motion,

v=drdt,a=dvdt,Fresultant=ma\mathbf{v}=\frac{d\mathbf{r}}{dt},\qquad \mathbf{a}=\frac{d\mathbf{v}}{dt},\qquad \mathbf{F}_{\mathrm{resultant}}=m\mathbf{a}

Integrate each component to recover v\mathbf{v} from a\mathbf{a}, or r\mathbf{r} from v\mathbf{v}, and use the supplied conditions.

Not in the formula booklet: learn it.

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