Chapter 3 ยท Motion in a Plane

3.1 Scalars, Vectors and Graphical Vector Addition

Some quantities are fixed by a number and a unit; others also need a direction and combine by the triangle rule. Vectors, multiplying them by numbers, and adding and subtracting them by drawing.

A delivery drone leaves its depot, flies $3.0\ \text{km}$ due east, then $4.0\ \text{km}$ due north and drops a parcel. Its battery has paid for $7.0\ \text{km}$ of flying, yet the customer is only $5.0\ \text{km}$ from the depot โ€” and to fly home the drone must know not just "$5.0\ \text{km}$" but "$5.0\ \text{km}$ in which direction". In the same way, two friends pulling a stuck car with $300\ \text{N}$ each, along ropes that make an angle, do not apply $600\ \text{N}$ to it. Adding $3$ and $4$ to get $5$, or $300$ and $300$ to get less than $600$, is not bad arithmetic: these quantities carry a direction, and quantities with direction add by a different rule.

In Chapter 2 a direction along a line could be handled by a plus or minus sign. In a plane (or in space) there are infinitely many directions, and we need a new kind of object โ€” the vector. This section defines vectors, shows how to stretch, shrink and reverse them by multiplying by a number, and how to add and subtract them by drawing: the triangle and parallelogram laws. Section 3.2 then turns the drawings into calculations.

Scalars and vectors

Some quantities are completely described by a number with a unit: the mass of a bag ($5.0\ \text{kg}$), a time interval ($12\ \text{s}$), a temperature ($300\ \text{K}$). Others are not: "a force of $20\ \text{N}$" does not tell you whether the box will slide left or right.

๐Ÿ“˜ Definition 3.1 โ€” Scalar and vector quantities

A scalar is a quantity specified completely by a single number and a unit. Scalars combine by ordinary arithmetic. Examples: mass, time, path length, speed, volume, density, temperature, work, energy, power.

A vector is a quantity that has both a magnitude and a direction and that adds according to the triangle law (Result 3.2 below). Examples: displacement, velocity, acceleration, force, momentum.

We print a vector in bold, $\mathbf{A}$ (by hand, write $\vec A$). Its magnitude is written $|\mathbf{A}|$ or simply $A$; a magnitude is never negative and carries the unit of the quantity. A vector is drawn as an arrow whose length (to a chosen scale) is its magnitude and whose arrowhead shows its direction.

The last condition in the definition matters. An electric current in a wire has a magnitude and a sense along the wire, but currents meeting at a junction simply add as numbers, whatever the angle between the wires. Current therefore does not obey the triangle law, and it is a scalar.

๐Ÿ“˜ Definition 3.2 โ€” Position vector and displacement vector

Choose an origin $O$ in a frame of reference. The position vector $\mathbf{r}$ of a point $P$ is the arrow drawn from $O$ to $P$; its magnitude is the distance $OP$.

If an object moves from $P$ (position vector $\mathbf{r}$) to $P'$ (position vector $\mathbf{r}'$), its displacement vector $\Delta\mathbf{r}$ is the arrow from $P$ to $P'$. It depends only on the start and end points โ€” not on the path taken between them, and not on where the origin is placed.

๐Ÿ“˜ Definition 3.3 โ€” Equal vectors and the negative of a vector

Two vectors $\mathbf{A}$ and $\mathbf{B}$ are equal, $\mathbf{A}=\mathbf{B}$, if they have the same magnitude and the same direction. An arrow may therefore be slid anywhere parallel to itself without changing the vector it represents.

The negative of $\mathbf{A}$, written $-\mathbf{A}$, has the same magnitude as $\mathbf{A}$ and the opposite direction.

โœ๏ธ Example 1 โ€” Scalar or vector?
Classify each quantity, giving a reason: (a) the $400\ \text{m}$ run in one lap of a track; (b) the runner's displacement after exactly one lap; (c) the velocity of a cricket ball; (d) the energy stored in a phone battery; (e) the force of the wind on a sail; (f) a current of $2\ \text{A}$ in a wire.
(a) A path length: a scalar ($400\ \text{m}$, no direction needed).
(b) A vector โ€” and here it is the zero vector, because the runner ends where she started.
(c), (e) Velocity and force: vectors. "A $30\ \text{m/s}$ ball" is incomplete until we say where it is going.
(d) Energy: a scalar.
(f) A scalar: although a current has a sense along the wire, currents add as plain numbers at a junction, so current does not obey the triangle law.
โœ๏ธ Example 2 โ€” Position vectors on a town map
Take the origin at a railway station, with east along $+x$ and north along $+y$. A cafรฉ is $300\ \text{m}$ east and $400\ \text{m}$ north of the station; a library is $600\ \text{m}$ due east of it. (a) Find the magnitude of the cafรฉ's position vector. (b) Find the displacement of a student who walks from the cafรฉ to the library. (c) Does the answer to (b) change if the origin is moved to the library?
(a) The cafรฉ is at the corner of a right triangle with sides $300\ \text{m}$ and $400\ \text{m}$: $|\mathbf{r}_\text{cafรฉ}|=\sqrt{300^2+400^2}=500\ \text{m}$.
(b) From the cafรฉ to the library the student ends up $600-300=300\ \text{m}$ further east and $400\ \text{m}$ further south. Magnitude $\sqrt{300^2+400^2}=500\ \text{m}$; direction $\tan\alpha=\tfrac{400}{300}$, $\alpha=53^\circ$. So $\Delta\mathbf{r}=500\ \text{m}$ at $53^\circ$ south of east.
(c) No. With the origin at the library both position vectors change, but the arrow from cafรฉ to library is the same arrow.
Check Station โ†’ cafรฉ ($300\ \text{m}$ E, $400\ \text{m}$ N) followed by cafรฉ โ†’ library ($300\ \text{m}$ E, $400\ \text{m}$ S) lands $600\ \text{m}$ E, $0\ \text{m}$ N โ€” exactly at the library โœ“. (That $|\mathbf{r}_\text{cafรฉ}|$ and $|\Delta\mathbf{r}|$ are both $500\ \text{m}$ is a coincidence of these numbers.)

Multiplying a vector by a real number

Doubling a velocity should give a velocity in the same direction, twice as large; reversing it should flip the arrow. This is multiplication by a real number.

๐Ÿ“ Result 3.1 โ€” Multiplying a vector by a real number

For a real number $\lambda$ and a vector $\mathbf{A}$, the product $\lambda\mathbf{A}$ is a vector with magnitude $|\lambda|\,A$, pointing along $\mathbf{A}$ if $\lambda\gt0$ and opposite to $\mathbf{A}$ if $\lambda\lt0$. If $\lambda=0$, the product is the zero vector. In particular $(-1)\mathbf{A}=-\mathbf{A}$.

If $\lambda$ is itself a physical quantity, the product is a new quantity with new units and dimensions: a constant velocity $\mathbf{v}$ multiplied by a time $\Delta t$ gives a displacement $\Delta\mathbf{r}=\mathbf{v}\,\Delta t$; a mass times an acceleration gives a force, $\mathbf{F}=m\mathbf{a}$ (Chapter 4). A positive scalar factor never changes the direction.

โœ๏ธ Example 3 โ€” A cyclist's velocity
A cyclist of mass (with bicycle) $70\ \text{kg}$ rides with constant velocity $\mathbf{v}=8.0\ \text{m/s}$ due north. Describe (a) $3\mathbf{v}$, (b) $-0.5\,\mathbf{v}$, (c) her displacement in $25\ \text{s}$, (d) the vector $m\mathbf{v}$ (her momentum).
(a) $3\mathbf{v}$: $24\ \text{m/s}$ due north.
(b) $|-0.5|\times8.0=4.0$, and the negative factor reverses it: $4.0\ \text{m/s}$ due south.
(c) $\Delta\mathbf{r}=\mathbf{v}\,\Delta t$: $8.0\ \text{m/s}\times25\ \text{s}=200\ \text{m}$ due north. The unit changed from $\text{m/s}$ to $\text{m}$; the direction did not.
(d) $70\ \text{kg}\times8.0\ \text{m/s}=560\ \text{kg m/s}$ due north.

Adding vectors graphically

Walk $3.0\ \text{km}$ east and then $4.0\ \text{km}$ north. Your total displacement is the single arrow from where you started to where you finished. This "one step after the other" picture is the definition of vector addition.

๐Ÿ“ Result 3.2 โ€” Triangle law and parallelogram law of vector addition

Triangle (head-to-tail) law. To add $\mathbf{B}$ to $\mathbf{A}$, draw $\mathbf{A}$, then draw $\mathbf{B}$ with its tail at the head of $\mathbf{A}$. The resultant $\mathbf{R}=\mathbf{A}+\mathbf{B}$ is the arrow from the tail of $\mathbf{A}$ to the head of $\mathbf{B}$.

Parallelogram law. Equivalently, draw $\mathbf{A}$ and $\mathbf{B}$ from a common tail and complete the parallelogram. $\mathbf{R}=\mathbf{A}+\mathbf{B}$ is the diagonal starting at the common tail.

The two constructions give the same arrow, because the opposite sides of a parallelogram are equal vectors.

๐Ÿ“ Result 3.3 โ€” Properties of vector addition
  • Commutative: $\mathbf{A}+\mathbf{B}=\mathbf{B}+\mathbf{A}$ (the two triangles are the two halves of one parallelogram).
  • Associative: $(\mathbf{A}+\mathbf{B})+\mathbf{C}=\mathbf{A}+(\mathbf{B}+\mathbf{C})$, so we may write $\mathbf{A}+\mathbf{B}+\mathbf{C}$.
  • Polygon law: to add several vectors, place them head to tail in any order; the resultant runs from the first tail to the last head.
  • $\mathbf{A}+(-\mathbf{A})=\mathbf{0}$, and only vectors of the same kind (displacement with displacement, force with force) may be added.
๐Ÿ“˜ Definition 3.4 โ€” The zero (null) vector

The zero vector $\mathbf{0}$ has magnitude zero; its direction is not defined. It is what you get from $\mathbf{A}+(-\mathbf{A})$, from $0\,\mathbf{A}$, or from the displacement of a body that returns to its starting point. For every vector $\mathbf{A}$ and number $\lambda$: $\ \mathbf{A}+\mathbf{0}=\mathbf{A}$, $\ \lambda\mathbf{0}=\mathbf{0}$, $\ 0\,\mathbf{A}=\mathbf{0}$.

In general the resultant must be found by a scale drawing (or by the calculation of Section 3.2), but three cases follow at once from the triangle law. Let the angle between $\mathbf{A}$ and $\mathbf{B}$, drawn tail to tail, be $\theta$:

For every other angle $R$ lies between the extremes: $\ |A-B|\le|\mathbf{A}+\mathbf{B}|\le A+B$ โ€” no side of a triangle is longer than the other two together.

โœ๏ธ Example 4 โ€” Two ropes on a stuck cart
Two ropes pull a cart with forces of $6.0\ \text{N}$ and $8.0\ \text{N}$. Find the resultant when the ropes are (a) in the same direction, (b) in opposite directions, (c) at right angles. (d) Could the resultant ever be $1.0\ \text{N}$, or $15\ \text{N}$?
(a) $R=6.0+8.0=14\ \text{N}$ along both ropes.
(b) $R=8.0-6.0=2.0\ \text{N}$ along the $8.0\ \text{N}$ rope.
(c) $R=\sqrt{6.0^2+8.0^2}=10\ \text{N}$, at $\alpha$ to the $6.0\ \text{N}$ rope with $\tan\alpha=\tfrac{8.0}{6.0}$, $\alpha=53^\circ$.
(d) No. Every resultant satisfies $2.0\ \text{N}\le R\le14\ \text{N}$.
Check $10\ \text{N}$ lies between $2.0$ and $14\ \text{N}$ โœ“, and it is not $6.0+8.0$: magnitudes add only when the vectors point the same way.
โœ๏ธ Example 5 โ€” Adding by scale drawing
A walker moves $\mathbf{A}=4.0\ \text{m}$ due east and then $\mathbf{B}=3.0\ \text{m}$ at $60^\circ$ north of east. Find her resultant displacement by a scale drawing.
Scale Choose $1\ \text{cm}$ on paper $=1\ \text{m}$. Draw $\mathbf{A}$ as a $4.0\ \text{cm}$ arrow pointing right (east).
Head to tail At the head of $\mathbf{A}$, use a protractor to draw $\mathbf{B}$: $3.0\ \text{cm}$ at $60^\circ$ above the east direction.
Measure Join the tail of $\mathbf{A}$ to the head of $\mathbf{B}$. A careful drawing gives a line about $6.1\ \text{cm}$ long at about $25^\circ$ above east: $\mathbf{R}\approx6.1\ \text{m}$ at $25^\circ$ north of east.
Check $R$ lies between $4.0-3.0=1.0\ \text{m}$ and $4.0+3.0=7.0\ \text{m}$, and its direction lies between those of $\mathbf{A}$ ($0^\circ$) and $\mathbf{B}$ ($60^\circ$), closer to the longer vector โœ“. Section 3.2 calculates it exactly: $6.08\ \text{m}$ at $25.3^\circ$. Set these vectors in Interactive 1 to see it.
Interactive 1 โ€” Adding and Subtracting Vectors by Drawing

Drag the round handles at the heads of $\mathbf{A}$ (red) and $\mathbf{B}$ (blue), or use the sliders. Switch between the head-to-tail triangle, the parallelogram and the difference $\mathbf{A}-\mathbf{B}$. Press Play to swing $\mathbf{B}$ round and watch the resultant (purple) grow and shrink between $|A-B|$ and $A+B$. One grid square is one unit.

60ยฐ3.0
$\mathbf{A}$โ€”
$\mathbf{B}$โ€”
angle between themโ€”
resultantโ€”

Subtracting vectors

Subtraction is addition of the negative: $\mathbf{A}-\mathbf{B}=\mathbf{A}+(-\mathbf{B})$. Draw $\mathbf{A}$ and $\mathbf{B}$ from a common tail; then $\mathbf{A}-\mathbf{B}$ is the arrow from the head of $\mathbf{B}$ to the head of $\mathbf{A}$, because $\mathbf{B}+(\mathbf{A}-\mathbf{B})=\mathbf{A}$. The most important use in this chapter is a change of a vector, $\Delta\mathbf{v}=\mathbf{v}_2-\mathbf{v}_1$: final minus initial.

โœ๏ธ Example 6 โ€” Turning a corner at constant speed
A car travelling at $10\ \text{m/s}$ due east turns a corner and then travels at $10\ \text{m/s}$ due north. Find the change in its velocity. Has its velocity changed even though its speed has not?
Set up $\Delta\mathbf{v}=\mathbf{v}_2-\mathbf{v}_1=\mathbf{v}_2+(-\mathbf{v}_1)$: $10\ \text{m/s}$ north plus $10\ \text{m/s}$ west.
Magnitude The two are perpendicular: $|\Delta\mathbf{v}|=\sqrt{10^2+10^2}=14\ \text{m/s}$ (more precisely $14.1\ \text{m/s}$).
Direction Equal westward and northward parts: north-west, $45^\circ$ west of north.
Meaning Yes โ€” the velocity changed by $14\ \text{m/s}$ although the speed stayed $10\ \text{m/s}$. A change of direction alone is a change of velocity, so a car rounding a bend is accelerating. This is the key to circular motion in Section 3.4.
โš ๏ธ Common mistakes

1. Adding magnitudes. $|\mathbf{A}+\mathbf{B}|=A+B$ only when $\mathbf{A}$ and $\mathbf{B}$ point the same way. $3\ \text{km}$ east plus $4\ \text{km}$ north is $5\ \text{km}$, not $7\ \text{km}$.

2. Reversing a difference. With tails together, $\mathbf{A}-\mathbf{B}$ points to $\mathbf{A}$'s head, from $\mathbf{B}$'s head. $\mathbf{B}-\mathbf{A}=-(\mathbf{A}-\mathbf{B})$ points the other way.

3. Mixing kinds. A velocity cannot be added to a displacement, even though both are vectors โ€” first multiply the velocity by a time.

4. "It has a direction, so it is a vector." A quantity is a vector only if it also adds by the triangle law (current is the standard counter-example).

5. Confusing position and displacement. Position vectors depend on the choice of origin; displacements do not.

Interactive 2 โ€” The Polygon Law: Order Does Not Matter

An orienteer runs a course made of straight legs. Press Play to watch the legs being laid head to tail; the purple arrow is the displacement from the start so far, and the counter shows the path length. Press Shuffle order to run the same legs in a different order: the route changes, but the final displacement does not โ€” vector addition is commutative and associative.

legs completedโ€”
path length so farโ€”
displacement from startโ€”
order of legsโ€”

Exercises

I
Scalars and vectors. State whether each is a scalar or a vector, with a reason.
  • 1 the speed shown by a car's speedometer
  • 2 the momentum of a moving ball
  • 3 the work done in lifting a box
  • 4 the pressure of the air in a tyre
  • 5 the weight of a school bag
  • 6 the electric current in a wire
  • 7 the displacement of the tip of a clock's hour hand from noon to midnight
  • 8 Can two vectors of different magnitudes add up to the zero vector? Can three?
1 Scalar: speed is the magnitude of velocity, with no direction.
2 Vector: $\mathbf{p}=m\mathbf{v}$, a positive number times a vector, has the direction of $\mathbf{v}$.
3 Scalar: work is an energy transfer.
4 Scalar: pressure pushes equally in all directions; the force it exerts on a particular surface is a vector.
5 Vector: weight is a force, directed vertically down.
6 Scalar: currents add as numbers at a junction, whatever the angle between the wires, so current does not obey the triangle law.
7 The zero vector: after 12 hours the tip is back where it started.
8 Two: no โ€” $|\mathbf{A}+\mathbf{B}|\ge|A-B|\gt0$ when $A\ne B$. Three: yes, if they can form a closed triangle head to tail, e.g. $3$, $4$ and $5$ units.
II
Multiplying by numbers; position vectors. A runner has constant velocity $\mathbf{v}=15\ \text{m/s}$ due west.
  • 1 Describe $2\mathbf{v}$.
  • 2 Describe $-3\mathbf{v}$.
  • 3 Describe $\tfrac13\mathbf{v}$.
  • 4 Find the runner's displacement in $4.0\ \text{s}$.
  • 5 A $2.5\ \text{kg}$ trolley has acceleration $\mathbf{a}=4.0\ \text{m/s}^2$ towards the north-east. Describe the vector $m\mathbf{a}$ and give its unit.
  • 6 With the origin at a school gate, $x$ east and $y$ north, point $P$ is $120\ \text{m}$ east and $50\ \text{m}$ north of the gate and point $Q$ is $40\ \text{m}$ west and $70\ \text{m}$ north of it. Find $|\mathbf{r}_P|$ and the displacement from $P$ to $Q$.
  • 7 In part 6 the origin is moved to a library $100\ \text{m}$ east of the gate. Which of $\mathbf{r}_P$, $\mathbf{r}_Q$ and the displacement $P\to Q$ change?
1 $30\ \text{m/s}$ due west.
2 Magnitude $3\times15=45\ \text{m/s}$, reversed: $45\ \text{m/s}$ due east.
3 $5.0\ \text{m/s}$ due west.
4 $\Delta\mathbf{r}=\mathbf{v}\,\Delta t=15\times4.0=60\ \text{m}$ due west.
5 $2.5\times4.0=10\ \text{kg m/s}^2=10\ \text{N}$ towards the north-east (a force, Chapter 4).
6 $|\mathbf{r}_P|=\sqrt{120^2+50^2}=130\ \text{m}$. From $P$ to $Q$: $160\ \text{m}$ west and $20\ \text{m}$ north, so $|\Delta\mathbf{r}|=\sqrt{160^2+20^2}=161\ \text{m}$ at $\tan^{-1}\tfrac{20}{160}=7.1^\circ$ north of west.
7 $\mathbf{r}_P$ and $\mathbf{r}_Q$ change (they are measured from the new origin); the displacement $P\to Q$ does not.
III
Graphical addition and special cases.
  • 1 Forces of $5.0\ \text{N}$ and $12.0\ \text{N}$ act on a body. Find the largest and smallest possible resultants.
  • 2 Find the resultant of the forces in part 1 when they are perpendicular (magnitude and angle with the $5.0\ \text{N}$ force).
  • 3 Can forces of $4\ \text{N}$, $7\ \text{N}$ and $12\ \text{N}$ acting on one point have a zero resultant?
  • 4 Can forces of $6\ \text{N}$, $8\ \text{N}$ and $10\ \text{N}$? If so, what is the angle between the $6\ \text{N}$ and $8\ \text{N}$ forces?
  • 5 A girl walks $40\ \text{m}$ north, $30\ \text{m}$ east and then $40\ \text{m}$ south. Find her displacement and path length.
  • 6 A hiker walks $2.0\ \text{km}$ east, $3.0\ \text{km}$ north and then $6.0\ \text{km}$ west. Find her resultant displacement.
  • 7 Two vectors each of magnitude $A$ are perpendicular. Find their resultant.
  • 8 Two vectors each of magnitude $A$ act at an angle $\theta$. For what $\theta$ is the resultant (a) $2A$, (b) zero, (c) equal to $A$? (Hint for (c): what kind of triangle is it?)
  • 9 Using three displacement arrows of your choice, sketch $(\mathbf{A}+\mathbf{B})+\mathbf{C}$ and $\mathbf{A}+(\mathbf{B}+\mathbf{C})$ and explain why they end at the same point.
1 Largest $17.0\ \text{N}$ (same direction); smallest $7.0\ \text{N}$ (opposite).
2 $\sqrt{5.0^2+12.0^2}=13.0\ \text{N}$ at $\tan^{-1}\tfrac{12.0}{5.0}=67.4^\circ$ to the $5.0\ \text{N}$ force.
3 No: they would have to form a closed triangle, but $12\gt4+7=11$.
4 Yes: $6^2+8^2=10^2$, so they close a right triangle with the $6\ \text{N}$ and $8\ \text{N}$ sides at $90^\circ$; drawn tail to tail, the $6\ \text{N}$ and $8\ \text{N}$ forces are at $90^\circ$ and their resultant ($10\ \text{N}$) is cancelled by the third force.
5 The north and south legs cancel: displacement $30\ \text{m}$ due east; path length $40+30+40=110\ \text{m}$.
6 Net $4.0\ \text{km}$ west and $3.0\ \text{km}$ north: $5.0\ \text{km}$ at $\tan^{-1}\tfrac34=36.9^\circ$ north of west (path length $11\ \text{km}$).
7 $\sqrt{A^2+A^2}=\sqrt2\,A$, at $45^\circ$ to each vector.
8 (a) $0^\circ$; (b) $180^\circ$; (c) the triangle $\mathbf{A}$, $\mathbf{B}$, $\mathbf{R}$ has three equal sides, so it is equilateral; the angle inside the triangle between the sides is $60^\circ$, so the angle between the vectors drawn tail to tail is $180^\circ-60^\circ=120^\circ$.
9 Both are the polygon $\mathbf{A}$, then $\mathbf{B}$, then $\mathbf{C}$ laid head to tail; the brackets only change which diagonal you draw on the way. The final head is the same, so the resultants are equal (associative law).
IV
Subtraction and applications.
  • 1 A ball hits a wall at $6.0\ \text{m/s}$ and rebounds straight back at $4.0\ \text{m/s}$. Find the change in its velocity.
  • 2 A runner goes round a circular track at a steady $5.0\ \text{m/s}$. Find the magnitude of the change in her velocity after (a) a quarter lap, (b) half a lap, (c) a full lap.
  • 3 The minute hand of a clock is $12\ \text{cm}$ long. For its tip, find the displacement and the path length between 3:00 and 3:15, and between 3:00 and 3:30.
  • 4 $\mathbf{A}=5.0\ \text{m}$ due east and $\mathbf{B}=5.0\ \text{m}$ due north. Find $\mathbf{A}-\mathbf{B}$ and $\mathbf{B}-\mathbf{A}$.
  • 5 A delivery drone flies $1.2\ \text{km}$ north, $0.90\ \text{km}$ east and then $0.40\ \text{km}$ south. Find its displacement from the depot, the path length, and the straight-line flight that returns it to the depot.
  • 6 Explain, with a sketch, why $|\mathbf{A}-\mathbf{B}|$ is larger than $|\mathbf{A}+\mathbf{B}|$ when the angle between $\mathbf{A}$ and $\mathbf{B}$ is more than $90^\circ$.
1 Take "away from the wall" as positive: $\Delta v=(+4.0)-(-6.0)=+10.0\ \text{m/s}$, i.e. $10.0\ \text{m/s}$ directed away from the wall (not $2\ \text{m/s}$!).
2 (a) the two velocities are perpendicular: $\sqrt{5.0^2+5.0^2}=7.1\ \text{m/s}$; (b) opposite: $5.0+5.0=10\ \text{m/s}$; (c) zero โ€” the velocity is the same again.
3 3:00โ€“3:15: a quarter turn; displacement $12\sqrt2=17\ \text{cm}$ (from the "3" towards the "6"), path $\tfrac14(2\pi\times12)=18.8\ \text{cm}$. 3:00โ€“3:30: displacement $24\ \text{cm}$ (a diameter, straight down), path $37.7\ \text{cm}$.
4 $\mathbf{A}-\mathbf{B}=\mathbf{A}+(-\mathbf{B})$: $5.0\ \text{m}$ east plus $5.0\ \text{m}$ south $=7.1\ \text{m}$ towards the south-east. $\mathbf{B}-\mathbf{A}$: $7.1\ \text{m}$ towards the north-west.
5 Net $0.80\ \text{km}$ north and $0.90\ \text{km}$ east: $\sqrt{0.80^2+0.90^2}=1.2\ \text{km}$ at $\tan^{-1}\tfrac{0.80}{0.90}=41.6^\circ$ north of east. Path length $2.5\ \text{km}$. Return: $1.2\ \text{km}$ at $41.6^\circ$ south of west (the negative vector).
6 With tails together, $\mathbf{A}+\mathbf{B}$ and $\mathbf{A}-\mathbf{B}$ are the two diagonals of the parallelogram. When the angle at the common tail is obtuse, the diagonal joining the two heads ($\mathbf{A}-\mathbf{B}$) is the longer one; at exactly $90^\circ$ (a rectangle) the diagonals are equal.
๐ŸŽฏ Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board ยท 2 marks Distinguish between a scalar and a vector quantity, giving one example of each. Electric current has a direction; is it a vector? Give a reason.

Definitions A scalar has magnitude only (e.g. mass); a vector has magnitude and direction and adds by the triangle law (e.g. force). [1]
Current No: currents add algebraically at a junction regardless of the angle between the wires, so current does not obey the triangle law; it is a scalar. [1]
Exam tip "Has a direction" is not enough โ€” always quote the triangle law when asked why something is (or is not) a vector.

E2. NEET Two forces of $3\ \text{N}$ and $4\ \text{N}$ act on a particle. Their resultant cannot be (a) $1\ \text{N}$ (b) $3.5\ \text{N}$ (c) $7\ \text{N}$ (d) $7.5\ \text{N}$

Range $|4-3|\le R\le4+3$, i.e. $1\ \text{N}\le R\le7\ \text{N}$.
Answer (d) $7.5\ \text{N}$ lies outside the range. (a) and (c) are the extreme cases (opposite and parallel forces).
Exam tip The trap is (a) or (c): the end points are possible.

E3. JEE Main The minute hand of a wall clock is $6.0\ \text{cm}$ long. The magnitude of the displacement of its tip between 12:00 and 12:20 is (a) $6.0\ \text{cm}$ (b) $10.4\ \text{cm}$ (c) $12.6\ \text{cm}$ (d) $12\ \text{cm}$

Angle In 20 minutes the hand turns $\tfrac{20}{60}\times360^\circ=120^\circ$.
Chord The displacement is the chord of a $120^\circ$ arc: an isosceles triangle with two sides $6.0\ \text{cm}$ and apex angle $120^\circ$. Splitting it into two right triangles, the chord is $2\times6.0\sin60^\circ=10.4\ \text{cm}$.
Answer (b).
Exam tip (c) $=\tfrac13(2\pi\times6.0)$ is the path length (arc), and (d) the diameter (after 30 minutes). Displacement is always the straight chord.

E4. NEET JEE Main Two vectors of equal magnitude have a resultant equal in magnitude to either of them. The angle between the two vectors is (a) $60^\circ$ (b) $90^\circ$ (c) $120^\circ$ (d) $180^\circ$

Triangle law $\mathbf{A}$, $\mathbf{B}$ (head to tail) and $\mathbf{R}$ form a triangle with three equal sides: equilateral, all inside angles $60^\circ$.
Angle between Drawn tail to tail, the angle between $\mathbf{A}$ and $\mathbf{B}$ is the exterior angle, $180^\circ-60^\circ=120^\circ$.
Answer (c).
Exam tip (a) is the inside angle of the triangle โ€” the most common wrong answer. Section 3.2 gives the formula $R=2A\cos\tfrac\theta2$, which confirms $R=A$ at $\theta=120^\circ$.

E5. Board ยท 3 marks A car moving at $20\ \text{m/s}$ due east turns a corner and is moving at $20\ \text{m/s}$ due north $4.0\ \text{s}$ later. (a) Has its velocity changed? Explain. (b) Find the magnitude and direction of the change in velocity. (c) Find the magnitude of its average acceleration $\dfrac{\Delta\mathbf{v}}{\Delta t}$ over the turn.

(a) Yes: the speed is unchanged but the direction has changed, and velocity includes direction. [1]
(b) $\Delta\mathbf{v}=\mathbf{v}_2+(-\mathbf{v}_1)$ = $20\ \text{m/s}$ north plus $20\ \text{m/s}$ west: $20\sqrt2=28\ \text{m/s}$ towards the north-west. [1]
(c) $\dfrac{28.3\ \text{m/s}}{4.0\ \text{s}}=7.1\ \text{m/s}^2$, directed north-west. [1]
Exam tip Draw $\mathbf{v}_1$ and $\mathbf{v}_2$ from one point and join head of $\mathbf{v}_1$ to head of $\mathbf{v}_2$; the examiner wants to see that arrow.

E6. JEE Advanced (One or more correct) Non-zero vectors satisfy $\mathbf{A}+\mathbf{B}=\mathbf{C}$. Which of the following are always true? (A) $C\le A+B$ (B) $C\ge|A-B|$ (C) if $C=A+B$, then $\mathbf{A}$ and $\mathbf{B}$ point in the same direction (D) $C$ is at least as large as the smaller of $A$ and $B$

(A), (B) These are the triangle inequalities for the triangle $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ โ€” always true โœ“.
(C) Equality in (A) means the "triangle" is flat with $\mathbf{B}$ continuing straight on from $\mathbf{A}$: same direction โœ“.
(D) False: $\mathbf{A}=5$ units east and $\mathbf{B}=4$ units west give $C=1$ unit, smaller than both โœ—.
Answer (A), (B), (C).
Exam tip For "always true" options, hunt for one counter-example; nearly antiparallel vectors are the usual one.
๐Ÿ”ฌ Researchโ€“Teaching NexusResearch-tutored ยท optional, about 30 minutes

Research question. How much further than "as the crow flies" do people really travel in your town?

Why it matters. Transport planners compare cities with the detour index (or circuity): the path length of a real route divided by the magnitude of the straight-line displacement. A grid city, a hilly town and a town split by a river have very different values; the index affects travel times, fuel use, where ambulances and fire stations are placed, and how much walking or cycling people do. Delivery-drone companies quote the same ratio to explain why flying can beat driving.

What to do.

  1. Pick five everyday trips (home to school, to a market, to a hospitalโ€ฆ). On an online map, measure the straight-line distance (the magnitude of the displacement) and the shortest road or footpath distance for each.
  2. Compute the detour index for each trip. Which trips have the largest index, and what in the town's layout causes it?
  3. For one trip, split the road route into 4โ€“6 roughly straight legs and draw them head to tail to scale. Check that the polygon closes on the straight-line displacement.
  4. For a perfect square grid of streets, show that a trip $3$ blocks east and $4$ blocks north has a detour index of $\tfrac75=1.4$. Which direction of travel gives the worst index on a grid, and what is its value?

Tools: An online map with a "measure distance" tool, ruler and protractor.

Hand in: A table of trips with displacement, path length and detour index, your scale polygon, and a short paragraph suggesting one change (a footbridge, a cut-through) that would lower the index most.

Going further. Drones fly the displacement, not the road. If a drone and a delivery van have the same average speed, by what factor is the drone faster for each of your trips?

Instructor note: real road networks usually give indices of about $1.2$ to $1.5$; rivers, railways and hills push individual trips much higher. On a square grid a trip at $45^\circ$ to the streets is worst, with index $\tfrac{2}{\sqrt2}=\sqrt2\approx1.41$ (the $3$โ€“$4$ trip gives $1.4$). The point to draw out: path length is a scalar that adds legs' lengths, while displacement is a vector that adds by the polygon law.

๐Ÿงช Self-check quiz

0/5 correct

Q1. Which of these is a vector quantity?

Q2. Two forces of $7\ \text{N}$ and $3\ \text{N}$ act on a body. Which value is impossible for their resultant?

Q3. The vector $-2\mathbf{A}$ has

Q4. $\mathbf{A}$ and $\mathbf{B}$ are drawn from a common tail. The vector $\mathbf{A}-\mathbf{B}$ is the arrow

Q5. A cyclist rides half-way round a circular track of radius $r$. The magnitude of her displacement is

โ† Chapter 3 โ€” Motion in a Plane3.2 Resolution of Vectors and Analytical Addition โ†’