M1 2.1, M1 2.2 - Vectors in motion

M1 2.1, M1 2.2 - Vectors in motion

Describe directions and resultants with two-dimensional vectors, then use them for displacement, velocity and acceleration. M1 2.1 supplies vector geometry; M1 2.2 connects that geometry to motion over a stated time interval.

Magnitude direction and components

Two boats both travel at 5 m s−15\,\mathrm{m\,s^{-1}}, but one heads east and the other north. Do they have the same velocity? Their speeds match; their directions do not.

A scalar has magnitude only, such as time or mass. A vector has magnitude and direction, such as displacement, velocity, acceleration or force. Choose perpendicular axes first. Here i\mathbf i is a unit vector east and j\mathbf j a unit vector north. A unit vector has magnitude 11; a negative component points opposite its axis.

The displacement 3i+4j3\mathbf i+4\mathbf j metres means 3 m3\,\mathrm m east and 4 m4\,\mathrm m north. Its straight-line length is 5 m5\,\mathrm m, by Pythagoras.

Magnitude and components

For perpendicular unit vectors i,j\mathbf i,\mathbf j,

∣xi+yj∣=x2+y2.\lvert x\mathbf i+y\mathbf j\rvert=\sqrt{x^2+y^2}.

A vector of magnitude RR at angle θ\theta anticlockwise from the positive horizontal axis has components

Rcos⁡θ i+Rsin⁡θ j.R\cos\theta\,\mathbf i+R\sin\theta\,\mathbf j.

Check the quadrant before interpreting an inverse tangent. These vector results are not printed in the formula booklet: learn them.

[DIAGRAM: asset_name: Vector components and quadrant; asset_slug: m1-vector-components; caption: The component signs locate the vector before trigonometry finds its angle.; description: Two equal-scale coordinate diagrams show vectors three i plus four j and negative three i plus four j, both of magnitude five, with horizontal and vertical component legs and angles north of east and north of west.]
The component signs locate the vector before trigonometry finds its angle.

Worked example 1: north of east

Question. A displacement is (3i+4j) m(3\mathbf i+4\mathbf j)\,\mathrm m. Find its magnitude and direction.

Plan. Use Pythagoras for magnitude and a right triangle for the angle north of east.

R=32+42=5,θ=tan⁡−1(43)=53.130…∘.R=\sqrt{3^2+4^2}=5, \qquad \theta=\tan^{-1}\left(\frac43\right)=53.130\ldots^\circ.

Answer. 5 m5\,\mathrm m, at 53.1∘53.1^\circ north of east.

Check. 5cos⁡θ=35\cos\theta=3 and 5sin⁡θ=45\sin\theta=4, recovering both components.

Change only the horizontal component's sign. The magnitude stays the same, but the quadrant changes.

Worked example 2: north of west

Question. A displacement is (−3i+4j) m(-3\mathbf i+4\mathbf j)\,\mathrm m. Find its magnitude, its angle north of west and its bearing to the nearest degree.

Plan. Locate the north-west quadrant, then use the same component triangle.

R=(−3)2+42=5,θ=tan⁡−1(43)=53.130…∘.R=\sqrt{(-3)^2+4^2}=5, \qquad \theta=\tan^{-1}\left(\frac43\right)=53.130\ldots^\circ.

A bearing is measured clockwise from north, so here

bearing=270∘+θ=323.130…∘.\text{bearing}=270^\circ+\theta=323.130\ldots^\circ.

Answer. 5 m5\,\mathrm m, 53.1∘53.1^\circ north of west; bearing 323∘323^\circ.

Check. The bearing lies between 270∘270^\circ and 360∘360^\circ, as a north-west direction should.

An inverse-tangent result alone does not specify a vector direction.

Watch out

The magnitude x2+y2\sqrt{x^2+y^2} is nonnegative. It does not retain the component signs, so it cannot tell you the quadrant.

Use the new component ratio in the same quadrant.

Resultants and vector diagrams

A resultant is the single vector equivalent to adding the vectors, so add components before finding a magnitude.

Walking 3 m3\,\mathrm m east and then 4 m4\,\mathrm m north gives a displacement of magnitude 5 m5\,\mathrm m, not 7 m7\,\mathrm m. The distance walked is 7 m7\,\mathrm m. The displacement joins the original start to the final finish.

Vector addition

To add vectors in a diagram, place the tail of the second at the head of the first without changing its length or direction. The resultant joins the first tail to the final head.

(ai+bj)+(ci+dj)=(a+c)i+(b+d)j.(a\mathbf i+b\mathbf j)+(c\mathbf i+d\mathbf j) =(a+c)\mathbf i+(b+d)\mathbf j.

Subtracting a vector means adding its opposite.

[DIAGRAM: asset_name: Adding force vectors head to tail; asset_slug: m1-vector-resultant; caption: The resultant joins the first tail to the final head.; description: An exact equal-scale vector diagram adds seven i plus two j to negative three i plus j, obtaining four i plus three j. The second vector is translated to begin at the head of the first.]
The resultant joins the first tail to the final head.

Two nonzero vectors are parallel if one is a nonzero scalar multiple of the other. A positive multiplier preserves direction; a negative multiplier reverses it. “Parallel to” therefore does not necessarily mean “in the same direction as”.

Worked example 3: two force vectors

Question. Forces F1=(7i+2j) N\mathbf F_1=(7\mathbf i+2\mathbf j)\,\mathrm N and F2=(−3i+j) N\mathbf F_2=(-3\mathbf i+\mathbf j)\,\mathrm N act on a particle. Find the resultant and its magnitude.

Plan. Combine horizontal and vertical components separately, then use Pythagoras.

R=(7−3)i+(2+1)j=(4i+3j) N,\mathbf R=(7-3)\mathbf i+(2+1)\mathbf j =(4\mathbf i+3\mathbf j)\,\mathrm N, ∣R∣=42+32=5 N.\lvert\mathbf R\rvert=\sqrt{4^2+3^2}=5\,\mathrm N.

Answer. Resultant (4i+3j) N(4\mathbf i+3\mathbf j)\,\mathrm N, magnitude 5 N5\,\mathrm N.

Check. Subtract F2\mathbf F_2 from the resultant: (4+3)i+(3−1)j=7i+2j(4+3)\mathbf i+(3-1)\mathbf j=7\mathbf i+2\mathbf j, recovering F1\mathbf F_1.

Change only the sign of the second force: its arrow is reversed before adding.

Worked example 4: reversing one force

Question. Replace F2\mathbf F_2 above by −F2=(3i−j) N-\mathbf F_2=(3\mathbf i-\mathbf j)\,\mathrm N. Find the new resultant and its magnitude.

Plan. Add the reversed components, rather than subtracting magnitudes.

R=(7+3)i+(2−1)j=(10i+j) N,\mathbf R=(7+3)\mathbf i+(2-1)\mathbf j =(10\mathbf i+\mathbf j)\,\mathrm N, ∣R∣=102+12=101 N.\lvert\mathbf R\rvert=\sqrt{10^2+1^2}=\sqrt{101}\,\mathrm N.

Answer. (10i+j) N(10\mathbf i+\mathbf j)\,\mathrm N, magnitude 101 N\sqrt{101}\,\mathrm N.

Check. The new resultant minus the old is 6i−2j6\mathbf i-2\mathbf j, exactly −2F2-2\mathbf F_2 as required when replacing F2\mathbf F_2 by its opposite.

Forces are vectors, even when the question eventually asks for a positive magnitude.

Watch out

In general ∣F1+F2∣≠∣F1∣+∣F2∣\lvert\mathbf F_1+\mathbf F_2\rvert\ne\lvert\mathbf F_1\rvert+\lvert\mathbf F_2\rvert. Add the vectors first. Equal magnitudes alone also do not establish equal directions.

Try the same reversed-vector structure with fresh values.

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