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M1.2 - Vectors in mechanics

Syllabus
2019
Topic
M1.2
Level
AS

Learning objectives

M1.2.1Magnitude and directionMagnitude and direction of a Students may be required to resolve a vector into two vector.; Resultant of vectors may components or use a vector diagram.; Questions may be set also be required. involving the unit vectors i and j.M1.2.2Application of vectorsApplication of vectors to Use of displacements, velocities, change of displacement accelerations and forces in a plane. velocity = in the case of constant time change of velocity velocity, and of acceleration = in the time case of constant acceleration, will be required.

Read and combine vectors in a plane

A vector has both magnitude and direction. In a plane, write v=ai+bj\mathbf v=a\mathbf i+b\mathbf j, where the perpendicular unit vectors i\mathbf i and j\mathbf j define the positive coordinate directions. The signed components aa and bb say how much of the vector acts in each direction.

∣v∣=a2+b2|\mathbf v|=\sqrt{a^2+b^2}

A direction angle can be found from the component triangle, but the signs of aa and bb must determine the correct quadrant. For a bearing, measure clockwise from north and give a three-figure angle. A calculator value such as tan⁡−1(b/a)\tan^{-1}(b/a) gives a reference angle only; it does not by itself determine the direction.

To resolve a vector of magnitude VV at an angle θ\theta to the positive i\mathbf i direction, usev=(Vcos⁡θ)i+(Vsin⁡θ)j.\mathbf v=(V\cos\theta)\mathbf i+(V\sin\theta)\mathbf j.Change signs to match the actual quadrant. If the angle is measured from the j\mathbf j direction, the sine and cosine roles interchange.

R=v1+v2+⋯=(∑ak)i+(∑bk)j\mathbf R=\mathbf v_1+\mathbf v_2+\cdots=\left(\sum a_k\right)\mathbf i+\left(\sum b_k\right)\mathbf j

Forces 10i+j10\mathbf i+\mathbf j N and −15i+6.5j-15\mathbf i+6.5\mathbf j N have resultantR=−5i+7.5j N,∣R∣=(−5)2+7.52=5132 N.\mathbf R=-5\mathbf i+7.5\mathbf j\text{ N},\qquad |\mathbf R|=\sqrt{(-5)^2+7.5^2}=\frac{5\sqrt{13}}2\text{ N}.Its components place it north-west. The reference angle west of north satisfies tan⁡α=5/7.5\tan\alpha=5/7.5, so its bearing is 360∘−α≈326∘360^\circ-\alpha\approx326^\circ.

Add vector components, not magnitudes. Parallel vectors have proportional components; vectors in opposite directions have a negative proportionality factor. When an angle is requested between two directions, check whether the smaller angle, a directed angle or a bearing is required before choosing the final value.

Use vectors to model motion and force

Displacement, velocity, acceleration and force are vectors: each must keep its components, direction and units. The same component rules apply to all four quantities, but their meanings are different.

Quantity Vector relationship Meaning
Displacement from AA to BB AB→=rB−rA\overrightarrow{AB}=\mathbf r_B-\mathbf r_A Change of position; distance AB=∣AB→∣AB=|\overrightarrow{AB}|.
Constant velocity v=(r2−r1)/(t2−t1)\mathbf v=(\mathbf r_2-\mathbf r_1)/(t_2-t_1) Change of displacement per unit time.
Position at constant velocity r=r0+tv\mathbf r=\mathbf r_0+t\mathbf v Initial position plus displacement travelled.
Constant acceleration a=(v2−v1)/(t2−t1)\mathbf a=(\mathbf v_2-\mathbf v_1)/(t_2-t_1) Change of velocity per unit time.
Velocity at constant acceleration v=u+ta\mathbf v=\mathbf u+t\mathbf a Each velocity component changes linearly with time.
Resultant force F=∑Fk\mathbf F=\sum\mathbf F_k Component sum of all forces acting on the particle.

Use a consistent time unit before dividing or multiplying. For example, velocity may be in m s−1\mathrm{m\,s^{-1}} or km h−1\mathrm{km\,h^{-1}}, acceleration in m s−2\mathrm{m\,s^{-2}} or km h−2\mathrm{km\,h^{-2}}, and force in newtons. Speed is the magnitude ∣v∣|\mathbf v|; it is a scalar and has no direction.

A particle starts at r0=2i+5j\mathbf r_0=2\mathbf i+5\mathbf j m and moves with constant velocity v=3i−2j\mathbf v=3\mathbf i-2\mathbf j m s−1^{-1}. After 44 s,r=(2i+5j)+4(3i−2j)=14i−3j m.\mathbf r=(2\mathbf i+5\mathbf j)+4(3\mathbf i-2\mathbf j)=14\mathbf i-3\mathbf j\text{ m}.Its displacement is 12i−8j12\mathbf i-8\mathbf j m, while the distance from its starting point is 122+(−8)2=413\sqrt{12^2+(-8)^2}=4\sqrt{13} m.

For two moving particles, form one relative vector consistently:AB→(t)=rB(t)−rA(t).\overrightarrow{AB}(t)=\mathbf r_B(t)-\mathbf r_A(t).They meet only if both components are zero at the same time. Their separation is ∣AB→(t)∣|\overrightarrow{AB}(t)|; minimising its square gives the same closest time without an unnecessary square root.

If velocity changes from −i+4j-\mathbf i+4\mathbf j to 5i−8j5\mathbf i-8\mathbf j m s−1^{-1} in 33 s at constant acceleration, thena=(5i−8j)−(−i+4j)3=2i−4j m s−2.\mathbf a=\frac{(5\mathbf i-8\mathbf j)-(-\mathbf i+4\mathbf j)}3=2\mathbf i-4\mathbf j\text{ m s}^{-2}.Subtraction order fixes the direction of the change.

Position, displacement and distance are not interchangeable; velocity and speed are not interchangeable. Never divide one vector by another. Equality or parallelism of vectors must be checked component by component, and a collision requires the same position at the same time—not merely equal speeds or directions.