Q3 - Vector constant-acceleration motion

Q3 - Vector constant-acceleration motion

Extend constant-acceleration equations to two dimensions, calculate speed and direction from velocity components, and solve conditions that must hold at the same time. You will distinguish where a particle is from the direction in which it is moving, then recover a motion model from observations.

Position and velocity component equations

Can a particle be north of its starting point while moving due east? Use east/north axes, with distances in metres and time in seconds. Start it at (2,−1)(2,-1), give it initial velocity 3i+4j3\mathbf i+4\mathbf j and constant acceleration 2i−2j2\mathbf i-2\mathbf j. After 22 seconds its northward velocity has disappeared, but its northward displacement has not. Two component equations will tell us exactly where it is.

You already know

For constant acceleration along a line, v=u+atv=u+at and s=ut+12at2s=ut+\frac12at^2. A position vector locates a point from the origin; displacement is final position minus initial position. Perpendicular components add as vectors.

Let i\mathbf i and j\mathbf j be fixed perpendicular unit vectors in the stated positive directions. They have magnitude 11 and no physical units. You may write a vector as xi+yjx\mathbf i+y\mathbf j or (xy)\begin{pmatrix}x\\y\end{pmatrix}. The third Cartesian unit vector k\mathbf k is unnecessary for this two-dimensional motion.

A constant acceleration vector has constant components. Apply a straight-line equation separately in each direction, using the same elapsed time. Define r0\mathbf r_0 as initial position, r\mathbf r as position and s=r−r0\mathbf s=\mathbf r-\mathbf r_0 as displacement. This position–displacement relation is not printed in the formula booklet: learn it.

Constant vector acceleration

For constant a\mathbf a, initial velocity u\mathbf u and elapsed time tt,

v=u+tas=tu+12t2as=12t(u+v)s=tv−12t2a\begin{aligned} \mathbf v&=\mathbf u+t\mathbf a \\ \mathbf s&=t\mathbf u+\tfrac12t^2\mathbf a \\ \mathbf s&=\tfrac12t(\mathbf u+\mathbf v) \\ \mathbf s&=t\mathbf v-\tfrac12t^2\mathbf a \end{aligned}

In the formula booklet.

Position and displacement are measured in metres, velocity in m s−1\mathrm{m\,s^{-1}} and acceleration in m s−2\mathrm{m\,s^{-2}}. Adding r0\mathbf r_0 to a displacement equation gives the corresponding position equation. Constant acceleration need not produce a straight path: the two velocity components can change in different ways.

Worked example 1: three different vectors

Question. A particle has r0=2i−j\mathbf r_0=2\mathbf i-\mathbf j metres, u=3i+4j m s−1\mathbf u=3\mathbf i+4\mathbf j\,\mathrm{m\,s^{-1}} and constant a=2i−2j m s−2\mathbf a=2\mathbf i-2\mathbf j\,\mathrm{m\,s^{-2}}. Find velocity, displacement and position after 22 seconds.

Plan. Use the vector velocity and displacement equations, then add the initial position.

v(2)=(3+2×2)i+(4−2×2)jvelocity components=7is(2)=(3×2+12×2×22)i+(4×2−12×2×22)jdisplacement components=10i+4jr(2)=(2i−j)+(10i+4j)add initial position=12i+3j\begin{aligned} \mathbf v(2)&=(3+2\times2)\mathbf i+(4-2\times2)\mathbf j && \text{velocity components} \\ &=7\mathbf i \\ \mathbf s(2)&=(3\times2+\tfrac12\times2\times2^2)\mathbf i \\ &\quad +(4\times2-\tfrac12\times2\times2^2)\mathbf j && \text{displacement components} \\ &=10\mathbf i+4\mathbf j \\ \mathbf r(2)&=(2\mathbf i-\mathbf j)+(10\mathbf i+4\mathbf j) && \text{add initial position} \\ &=12\mathbf i+3\mathbf j \end{aligned}

Why add r0\mathbf r_0 only at the last step? The motion equation finds change in position, whereas r\mathbf r measures position from the chosen origin.

Answer. Velocity is 7i m s−17\mathbf i\,\mathrm{m\,s^{-1}}, displacement is 10i+4j10\mathbf i+4\mathbf j metres and position is 12i+3j12\mathbf i+3\mathbf j metres. It is north of its start while moving due east.

Check. Average velocity is 12[(3,4)+(7,0)]=(5,2)\tfrac12[(3,4)+(7,0)]=(5,2), giving displacement 2(5,2)=(10,4)2(5,2)=(10,4).

The path x=2+3t+t2x=2+3t+t^2, y=−1+4t−t2y=-1+4t-t^2 bends even though acceleration is constant. The arrow below is tangent to the path, rather than pointing from the origin to the particle.

Exact path with initial position, position at two seconds and its horizontal tangent direction

Worked example 2 changes the elapsed time only. The northward velocity now becomes negative, so the same path is being followed downwards.

Worked example 2: continuing past the horizontal tangent

Question. For the same particle, find velocity, displacement and position after 33 seconds.

Plan. Substitute t=3t=3 into the same component equations and keep the initial position separate.

v(3)=(3+6)i+(4−6)jvelocity equation=9i−2js(3)=(9+9)i+(12−9)jdisplacement equation=18i+3jr(3)=(2+18)i+(−1+3)jinitial position plus change=20i+2j\begin{aligned} \mathbf v(3)&=(3+6)\mathbf i+(4-6)\mathbf j && \text{velocity equation} \\ &=9\mathbf i-2\mathbf j \\ \mathbf s(3)&=(9+9)\mathbf i+(12-9)\mathbf j && \text{displacement equation} \\ &=18\mathbf i+3\mathbf j \\ \mathbf r(3)&=(2+18)\mathbf i+(-1+3)\mathbf j && \text{initial position plus change} \\ &=20\mathbf i+2\mathbf j \end{aligned}

Why can the displacement have a positive northward component while velocity has a negative one? Displacement records the whole journey so far; velocity describes motion at this instant.

Answer. Velocity is 9i−2j m s−19\mathbf i-2\mathbf j\,\mathrm{m\,s^{-1}}, displacement is 18i+3j18\mathbf i+3\mathbf j metres and position is 20i+2j20\mathbf i+2\mathbf j metres.

Check. The average-velocity relation gives 32[(3,4)+(9,−2)]=(18,3)\tfrac32[(3,4)+(9,-2)]=(18,3), agreeing with the displacement.

Both examples use legitimate vector sums. Squaring a vector as if it were a scalar would introduce an operation that these equations do not define.

Watch out

Wrong: v2=u2+2as\mathbf v^2=\mathbf u^2+2\mathbf a\mathbf s. There is no such vector equation in this course or the booklet. Right: use the four vector equations above, or apply vx2=ux2+2axsxv_x^2=u_x^2+2a_xs_x and its vertical counterpart to signed scalar components.

Try one more elapsed time. State all three vectors so that position and displacement cannot be confused.

Speed, direction and bearings

Velocity components determine both speed and direction; a position vector or a single velocity component cannot replace them.

If velocity is −6i+8j-6\mathbf i+8\mathbf j, the particle moves west and north. Pythagoras gives speed 1010, while a trigonometric ratio gives an angle. Keeping those two jobs separate avoids negative speeds and ambiguous directions.

Magnitude and direction of velocity

For fixed perpendicular unit vectors and v=vxi+vyj\mathbf v=v_x\mathbf i+v_y\mathbf j,

∣v∣=vx2+vy2|\mathbf v|=\sqrt{v_x^2+v_y^2}

When both components are nonzero, the acute reference angle α\alpha to a horizontal line satisfies

tan⁡α=∣vy∣∣vx∣\tan\alpha=\frac{|v_y|}{|v_x|}

Use the component signs to choose the quadrant. Not in the formula booklet: learn it.

If vx=0v_x=0 but vy≠0v_y\ne0, the direction is vertical. If both components vanish, the particle is at rest and its velocity has no direction. With i\mathbf i east and j\mathbf j north, a bearing is measured clockwise from north and written using three figures, such as 045∘045^\circ. This convention is not supplied in the booklet: learn it.

Worked example 3: moving north-west

Question. A particle has velocity −6i+8j m s−1-6\mathbf i+8\mathbf j\,\mathrm{m\,s^{-1}}, with east/north axes. Find its speed and bearing to the nearest degree.

Plan. Find the magnitude, then measure the north-west direction clockwise from north.

∣v∣=(−6)2+82=10Pythagorasα=tan⁡−1(6/8)≈36.87∘west of northbearing=360∘−α≈323.13∘clockwise from north\begin{aligned} |\mathbf v|&=\sqrt{(-6)^2+8^2}=10 && \text{Pythagoras} \\ \alpha&=\tan^{-1}(6/8)\approx36.87^\circ && \text{west of north} \\ \text{bearing}&=360^\circ-\alpha\approx323.13^\circ && \text{clockwise from north} \end{aligned}

Why use 6/86/8 here? The angle is measured from the northward component, so the westward component is opposite that angle.

Answer. Speed is 10 m s−110\,\mathrm{m\,s^{-1}} and bearing is 323∘323^\circ.

Check. The direction lies between west, 270∘270^\circ, and north, 360∘360^\circ, as the signs require.

Worked example 4 changes the sign of the eastward component only. The speed is unchanged, but the compass quadrant changes.

Worked example 4: the reflected direction

Question. Find the speed and bearing of velocity 6i+8j m s−16\mathbf i+8\mathbf j\,\mathrm{m\,s^{-1}}, using east/north axes. Give the bearing to the nearest degree.

Plan. Keep the same magnitude calculation and measure the angle east of north.

∣v∣=62+82=10magnitudeα=tan⁡−1(6/8)≈36.87∘east of northbearing≈037∘three-figure convention\begin{aligned} |\mathbf v|&=\sqrt{6^2+8^2}=10 && \text{magnitude} \\ \alpha&=\tan^{-1}(6/8)\approx36.87^\circ && \text{east of north} \\ \text{bearing}&\approx037^\circ && \text{three-figure convention} \end{aligned}

Why does the earlier subtraction disappear? The new velocity points clockwise from north into the north-east quadrant.

Answer. Speed is 10 m s−110\,\mathrm{m\,s^{-1}} and bearing is 037∘037^\circ.

Check. Reflecting north-west to north-east preserves length and gives bearings whose unrounded values sum to 360∘360^\circ.

Under constant vector acceleration, both speed and direction may change. A magnitude calculation removes direction information; it does not turn two-dimensional motion into one-dimensional motion.

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