Use vectors in geometry, relative motion and force models

Begin a geometric argument with definitions

Choose an origin and name each position vector. A displacement is a difference of positions. To show two nonzero segments are parallel, exhibit a scalar multiple; to show equal midpoints, equate their position averages. A numeric example illustrates a claim but cannot prove it for arbitrary points

Divide a segment with weighted positions

For distinct endpoints A,B and positive m,n, the internal point divides the segment in the ratio AP:PB=m:n. The coefficient of b records the fraction travelled from A. Equal ratio terms give the midpoint

For A=(1,−2),B=(5,0),find the x-coordinate of midpoint P.

For A=(1,−2),C=(−1,4),find the y-coordinate of midpoint Q.

Prove three geometric results, ungraded

For an arbitrary nondegenerate triangle, prove the midpoint theorem and that its medians meet at a point dividing each median in ratio 2:1. For a nondegenerate quadrilateral, prove the equivalence between being a parallelogram and having diagonals with the same midpoint. State the position-vector definitions. These proofs are ungraded

Model: midpoint theorem

If p=(a+b)/2 and q=(a+c)/2, then q−p=(c−b)/2. Since B≠C, PQ is parallel to BC and has half its length. The calculation holds for arbitrary triangle positions, not just the displayed coordinate example

Model: diagonals of a parallelogram

For consecutive vertices A,B,C,D, a parallelogram gives b−a=c−d, hence a+c=b+d. Dividing by 2 says the diagonal midpoints agree. Conversely equal midpoints give a+c=b+d, so b−a=c−d and d−a=c−b: opposite directed sides are equal. Under the nondegeneracy hypothesis this is a parallelogram

Model: the common median point

Set g=(a+b+c)/3. If M is the midpoint of BC, then g−a=(2/3)(m−a). Thus G lies two-thirds of the way from A to M. Applying the same calculation from B and C shows all three medians contain G and AG:GM=2:1. Nondegeneracy ensures these median segments are nonzero

For A=(1,−2),B=(5,0),C=(−1,4),find the centroid x-coordinate

State a frame before combining velocities

All position and velocity components must refer to compatible axes,units and times. For constant velocity in a fixed frame, r(t)=r₀+tv. The velocity of A relative to B is vA−vB; the relative position from B to A is rA−rB. Speed is a magnitude,not a signed component

If vA=(4,1) and vB=(1,−2) m/s,find the x-component of A relative to B.

With r₀=(−1,2) m and constant v=(3,−1) m/s,find x after 4 s.

A crossing path is not necessarily a collision

Collision of ideal point particles requires rA(t)=rB(t) at the same allowed time. Solve both coordinate equations with one common t. A solution in the past is excluded when t≥0. Equal velocities with distinct starting positions never meet; equal positions and equal velocities remain coincident

A follows rA(t)=(t,0) and B follows rB(t)=(2,t−1),for t≥0. Do they collide?

Closest approach uses relative motion

Minimise the squared separation for nonzero relative velocity. The displayed rule gives the unrestricted time before applying the future-time restriction. For t≥0, use the larger of that value and 0. If w=0,separation is constant and the earliest minimising time is0.

For r₀=(6,2) m and w=(−2,0) m/s,find the future time of closest approach

For r₀=(6,2) m and w=(−2,0) m/s over t≥0, find the squared minimum distance in square metres

Justify the relative-motion test, ungraded

Derive the relative position formula and closest-approach rule without assuming the two paths intersect. Include the zero-relative-velocity case and the future-time restriction. Explain why a zero minimum distance is required for a future point-particle collision. This reasoning is ungraded

Model: subtract positions and complete a square

Subtract the initial position and velocity of B from those of A to obtain the relative position. For nonzero relative velocity, complete the square as shown. The square is minimised at the stated unrestricted time; if that time is negative,the minimum over t≥0 occurs at 0. If w=0,all distances are the same. Future collision requires the minimum to be zero,with the same time used in both coordinates

Model forces acting on one particle

Choose the object and list every external force acting on it. A force component is not an additional force. The resultant is the vector sum. For constant positive mass in an inertial frame,Newton’s second law is Fnet=ma. Translational equilibrium means Fnet=0; a particle is in static equilibrium only if it is also at rest. A rigid body additionally needs zero resultant moment

If all forces are F₁=(3,−2) N, F₂=(−1,5) N and F₃, find the y-component of F₃ in newtons for zero resultant

A particle is already moving and its resultant force becomes zero. What follows in this ideal model?

Cable tensions must be feasible

A taut cable pulls along its unit direction away from the particle. For two cables with directions eL,eR and known external force F,equilibrium requires TL eL+TR eR+F=0. Solve both components and require TL,TR≥0. Negative tension would require a cable to push,so the assumed configuration is infeasible

Let eL=(−3/5,4/5),eR=(3/5,4/5),F=(6/5,−8) N. Find TL

Let eL=(−3/5,4/5), eR=(3/5,4/5) and F=(6/5,−8) N. Find TR for equilibrium

Constant net force gives constant acceleration

With constant mass m>0 and constant net force F,a=F/m. Over time t,velocity is v=u+ta and displacement is s=tu+½t²a. These are vector equations applied component by component. The supplied net force already includes gravity,resistance and other forces; do not add them again

A 2 kg particle has net force(4,−2) N. Find ay in m/s²

A 2 kg particle has net force (4,−2) N and initial velocity (1,3) m/s. Find vx after 2 s.

A 2 kg particle has net force (4,−2) N and initial velocity (1,3) m/s. Find its y-displacement over 2 s.

Does a force direction always equal the particle’s current velocity direction?

Derive and check the constant-force calculation, ungraded

Explain why constant mass and constant net force give constant acceleration. Derive displacement using the average velocity over the interval. Then verify the calculation for m=2 kg, F=(4,−2) N, u=(1,3) m/s and t=2 s component by component. State why displacement magnitude need not equal travelled distance. The derivation is ungraded

Model: average a linear velocity change

Fnet=ma gives constant a=Fnet/m. Velocity changes linearly from u to v=u+ta,so its average is(u+v)/2. Hence s=t(u+v)/2=tu+½t²a. For m=2,F=(4,−2),u=(1,3),t=2,the results are a=(2,−1),v=(5,1),s=(6,4). These follow in each component. The magnitude|s| is endpoint separation; the path can curve or reverse,so it need not equal the distance travelled

Construct and assess a physical model, ungraded

Choose either the two-cable equilibrium with eL=(−3/5,4/5), eR=(3/5,4/5) and F=(6/5,−8) N, or a 2 kg particle with net force (4,−2) N, initial velocity (1,3) m/s and a 2 s interval. State the frame, object, forces, units and assumptions; draw a labelled free-body diagram; calculate a prediction and describe evidence that would test it. Explain what changes if a cable goes slack or the net force varies. This modelling task is ungraded

Model: assumptions make the calculation conditional

For the cable case,treat the object as a particle at rest with two taut massless cables,known unit directions and all other forces represented once by F. With eL=(−3/5,4/5), eR=(3/5,4/5) and F=(6/5,−8) N, predict TL=6 N and TR=4 N, then check the component resultant is zero. Compare measured directions and tensions with the prediction; slack cables invalidate the positive-tension assumption. For motion,use constant mass,a fixed inertial frame and a constant measured net force. Track position/velocity at several times and compare with the predicted components. A varying force needs a revised model; a correct substitution alone does not establish physical validity