Kinematics

A car covers the first half of the distance at 40 km/h and the second half at 60 km/h. Find its average speed.

Almost everyone answers 50. It is 48.

Let each half be $d$. Time $=\frac{d}{40}+\frac{d}{60}=\frac{d}{24}$, so average speed $=\frac{2d}{d/24}=48$ km/h.

The car spends more time at the slower speed, so the slower speed gets more weight. Average speed is always weighted by time, never by distance.

If instead it travelled at those speeds for equal times, the answer would be 50 — the arithmetic mean.

That is this chapter in miniature: identify what is constant and what is being averaged over, and the method picks itself.

1. Position, distance, displacement

Position — a vector from a chosen origin. The origin is your free choice; nothing physical depends on it.

Displacement — change in position. A vector from start to finish. Ignores the route.

Distance — actual path length. A scalar. Can never decrease.

Example: 5 m east then 3 m west $\Rightarrow$ distance 8 m, displacement 2 m east.

Always true: distance $\ge$ |displacement|, with equality only if the motion never reverses. Use it as a check on any answer.

Closed path $\Rightarrow$ displacement zero, distance non-zero. A runner completing a lap has zero average velocity but non-zero average speed.

2. Scalars and vectors

Scalar — magnitude only: distance, speed, time, mass, work. Vector — magnitude and direction, obeying the triangle law of addition: displacement, velocity, acceleration, force.

The direction requirement is not enough on its own. Electric current has a direction and is still a scalar, because currents at a junction add arithmetically, not by the triangle law.

Unit vector — direction stripped of size: $\hat{A}=\frac{\vec{A}}{|\vec{A}|}$, so $\vec{A}=|\vec{A}|\hat{A}$. The Cartesian trio $\hat{i},\hat{j},\hat{k}$ are mutually perpendicular unit vectors.

Addition. Place them tail to tail and complete the parallelogram, or nose to tail and close the triangle. Both give the same resultant:

$R=\sqrt{A^2+B^2+2AB\cos \theta },\tan \alpha =\frac{B\sin \theta }{A+B\cos \theta }$

Reading the extremes off that formula is worth more than memorising it: $\theta =0$ gives $A+B$, $\theta =180°$ gives $|A-B|$, and $\theta =90°$ gives $\sqrt{A^2+B^2}$. Any resultant must lie between $|A-B|$ and $A+B$ — a one-line check on every vector answer.

Subtraction is addition of the reverse: $\vec{A}-\vec{B}=\vec{A}+(-\vec{B})$. This is the whole content of relative velocity in section 8.

Resolution is addition run backwards — replacing one vector by two perpendicular ones that can be handled separately. Choosing those two directions well is most of the skill in mechanics.

Two products, and they are not interchangeable:

In components, $\vec{A}\cdot \vec{B}=A_x{B}_x+A_y{B}_y+A_z{B}_z$, and the cross product is the $3\times 3$ determinant with $\hat{i},\hat{j},\hat{k}$ in the top row.

Illustration 1

Two vectors of magnitudes 6 and 8 act at $60°$ to each other. Find the resultant's magnitude and its direction.

$R=\sqrt{36+64+2(6)(8)\cos 60°}=\sqrt{100+48}=\sqrt{148}=12.2$

$\tan \alpha =\frac{8\sin 60°}{6+8\cos 60°}=\frac{6.93}{10}=0.693\Rightarrow \alpha =34.7°$ from the 6-unit vector.

Bounds check: the answer must lie between $|8-6|=2$ and $8+6=14$. It does, and it sits nearer the top because $60°$ is closer to parallel than to antiparallel.

Illustration 2

For $\vec{A}=2\hat{i}+3\hat{j}-\hat{k}$ and $\vec{B}=\hat{i}-2\hat{j}+2\hat{k}$, find $\vec{A}\cdot \vec{B}$, $\vec{A}\times \vec{B}$ and the angle between them.

$\vec{A}\cdot \vec{B}=2(1)+3(-2)+(-1)(2)=-6$

$|\vec{A}|=\sqrt{14}=3.74$, $|\vec{B}|=\sqrt{9}=3$

$\cos \theta =\frac{-6}{3.74(3)}=-0.535\Rightarrow \theta =122.3°$

The negative dot product already told us the angle is obtuse, before any arccosine.

$\vec{A}\times \vec{B}=(3\cdot 2-(-1)(-2))\hat{i}-(2\cdot 2-(-1)(1))\hat{j}+(2(-2)-3(1))\hat{k}=4\hat{i}-5\hat{j}-7\hat{k}$

$|\vec{A}\times \vec{B}|=\sqrt{16+25+49}=\sqrt{90}=9.49$

Cross-check the angle: $|\vec{A}||\vec{B}|\sin \theta =11.22\sin 122.3°=11.22(0.845)=9.49$. The two products agree, which is the standard way to catch a slip in either.

3. Velocity and speed

${\vec{v}}_{avg}=\frac{\Delta \vec{r}}{\Delta t},{\text{speed}}_{avg}=\frac{\text{total distance}}{\Delta t}$

$\vec{v}=\frac{d\vec{r}}{dt}$

Instantaneous speed $=$ |instantaneous velocity|, always. Only the averages can differ.

Trap. Which mean applies depends entirely on whether the halves are equal in distance or in time. Examiners set both in the same paper.

4. Acceleration

$\vec{a}=\frac{d\vec{v}}{dt}=\frac{d^2\vec{r}}{d{t}^2}$

Velocity is a vector, so acceleration arises from a change in magnitude or direction.

• $\vec{a}\parallel \vec{v}$ — speeding up
• $\vec{a}$ antiparallel to $\vec{v}$ — slowing down
• $\vec{a}\perp \vec{v}$ — direction changes, speed does not

Trap. Uniform circular motion is accelerated. Constant speed, changing direction.

"Deceleration" is not a separate concept — it just means $\vec{a}$ opposes $\vec{v}$. Writing $a$ negative without checking the direction of motion is where the sign errors come from.

5. Constant acceleration

Integrate $a=\frac{dv}{dt}$ once:

$v=u+at$

Integrate again for position:

$s=ut+\frac{1}{2}a{t}^2$

Eliminate $t$ between them:

$v^2=u^2+2as$

Displacement in the $n$-th second alone:

$s_n=u+\frac{a}{2}(2n-1)$

All four require constant $a$. Applying them to varying acceleration is the single biggest mark-loser in this chapter. If $a$ varies, integrate.

Pick a positive direction once and keep it. Most sign errors come from flipping convention halfway — typically making up positive on the rise and down positive on the fall.

When acceleration is not constant

Go back to the definitions and integrate. Nothing else is safe.

$v=u+\int adt,x=x_0+\int vdt$

If $a$ is given as a function of position rather than time, use the chain-rule form instead:

$a=v\frac{dv}{dx}$

Illustration 3

A particle starts from rest at the origin with $a=6t{\text{m/s}}^2$. Find its velocity and position at $t=2\text{s}$.

$v={\int }_0^{t}6tdt=3t^2\Rightarrow v(2)=12\text{m/s}$

$x={\int }_0^{t}3t^{2}dt=t^3\Rightarrow x(2)=8\text{m}$

The trap, shown explicitly: substituting $a=6(2)=12$ into $v=u+at$ gives $24$ m/s — double the right answer, because that formula assumes the acceleration held at 12 for the whole two seconds when in fact it grew from zero. Averaging it instead, $\bar{a}=6$, happens to give 12 m/s here only because $a$ is linear in $t$.

Illustration 4

From rest, $a=2{\text{m/s}}^2$. Find the displacement in the 5th second.

$s_5=0+\frac{2}{2}(2\times 5-1)=9\text{m}$

Check directly: $s_{5\text{s}}=\frac{1}{2}(2)(25)=25$ m, $s_{4\text{s}}=\frac{1}{2}(2)(16)=16$ m, difference $=9$ m.

Note $s_n$ has units of metres but is a displacement per second interval — numerically the average velocity during that second. From rest the successive seconds give 1, 3, 5, 7, 9 m: an AP with common difference $a$.

6. Graphs

Area below the axis is negative. That is exactly how a $v$–$t$ graph separates the two: displacement is the signed area, distance is the total unsigned area.

• A position–time graph can never be vertical — two positions at one instant is impossible.
• Horizontal position–time $\Rightarrow$ at rest.
• Concave up $\Rightarrow$ positive acceleration; concave down $\Rightarrow$ negative.
• Straight-line $v$–$t$ $\Rightarrow$ constant acceleration, i.e. the condition for section 5.

7. Motion under gravity

Free fall near the surface: constant downward $g\approx 9.8{\text{m/s}}^2$, independent of mass, air resistance neglected.

Every constant-acceleration result applies with $a=-g$ if up is positive.

Thrown up with speed $u$:

The last follows from $v^2=u^2+2as$ with $s=0$.

Conceptual favourite. Ascent and descent times are equal only without air resistance. With drag, the descent takes longer.

Illustration 5

Ball thrown up at 20 m/s from a 25 m tower. $g=10$. Find the time to hit the ground and the impact speed.

Take up positive, origin at the throwing point $\Rightarrow$ ground is at $s=-25$, $u=+20$, $a=-10$.

$-25=20t-5t^2\Rightarrow t^2-4t-5=0\Rightarrow (t-5)(t+1)=0$

$t=5\text{s}$ (reject $t=-1$).

$v=20-10(5)=-30\text{m/s}$ $\Rightarrow$ speed 30 m/s downward.

Check with $v^2=u^2+2as$: $400+2(-10)(-25)=900\Rightarrow |v|=30$.

Note the ball is back at launch height at $t=4$ s moving at 20 m/s down, then covers the remaining 25 m in just 1 more second.

8. Relative velocity

${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$

Vector subtraction — componentwise or by triangle. Never subtract magnitudes in two dimensions.

Antisymmetry: ${\vec{v}}_{AB}=-{\vec{v}}_{BA}$. Each observer sees the other at the same speed, opposite direction.

River crossing. Shortest time: aim straight across — the crossing time depends only on the component perpendicular to the bank. Shortest path: aim upstream so the upstream component exactly cancels the current.

Rain and umbrella is the same mathematics: rain's velocity relative to you is ${\vec{v}}_{\text{rain}}-{\vec{v}}_{\text{you}}$, and the umbrella tilts along that.

Illustration 6

River 100 m wide, current 3 m/s, boat 5 m/s in still water. Find (a) shortest crossing time and the drift, (b) the steering direction to land directly opposite.

(a) Aim straight across $\Rightarrow$ full 5 m/s is perpendicular.

$t=\frac{100}{5}=20\text{s}$, drift $=3\times 20=60\text{m}$

(b) Cancel the current: the upstream component must be 3 m/s.

$\sin \theta =\frac{3}{5}\Rightarrow \theta =37°$ upstream of the perpendicular.

Effective crossing speed $=\sqrt{25-9}=4$ m/s, so $t=25$ s — slower, as the shortest path always is.

9. Projectile motion

The whole idea: horizontal and vertical motions are independent. Gravity acts only vertically, so $u_x$ is constant and the vertical motion is uniformly accelerated.

$u_x=u\cos \theta (\text{constant}),u_y=u\sin \theta$

$T=\frac{2u\sin \theta }{g},H=\frac{u^2{\sin }^2\theta }{2g},R=\frac{u^2\sin 2\theta }{g}$

• $R$ is maximum at $\theta =45°$, because $\sin 2\theta$ peaks at $2\theta =90°$.
• Two angles give the same range. $\sin 2\theta =\sin (180°-2\theta )$, so $\theta$ and $90°-\theta$ share a range — with different $T$ and different $H$.

Trap. At the highest point the velocity is horizontal, not zero. Only $u_y$ vanishes there.

Horizontal projection is the same treatment with $u_y=0$: the fall time depends only on the height, so a ball rolled off a table and a ball dropped beside it land together.

Illustration 7

A ball rolls off a table $20\text{m}$ high at $5\text{m/s}$. $g=10$. Find the flight time, the horizontal distance, and the impact speed and angle.

Vertical motion alone fixes the time, and $u_y=0$:

$20=\frac{1}{2}(10)t^2\Rightarrow t=2\text{s}$

Horizontal: $x=5(2)=10\text{m}$

At impact, $v_x=5$ (unchanged) and $v_y=10(2)=20$:

$v=\sqrt{25+400}=20.6\text{m/s},\tan \alpha =\frac{20}{5}=4\Rightarrow \alpha =76°$ below the horizontal

Read it back: the 2 s is exactly what a ball simply dropped from the same table would take. The horizontal launch bought distance, not time — which is the independence principle made visible.

Illustration 8

Launched at 20 m/s with range 20 m. $g=10$. Find the possible angles.

$20=\frac{400\sin 2\theta }{10}\Rightarrow \sin 2\theta =0.5\Rightarrow 2\theta =30°$ or $150°$

$\theta =15°$ or $75°$ — complementary, as expected.

The two trajectories differ sharply:

Same range, one flat and fast, one high and slow.

Illustration 9

A and B are 100 m apart on a straight road, B ahead. A moves at 20 m/s, B at 10 m/s, same direction. When does A catch B?

Work in B's frame: $v_{AB}=20-10=10$ m/s, and A must close 100 m.

$t=\frac{100}{10}=10\text{s}$

Position: A has travelled $20\times 10=200$ m from its start.

Switching to the relative frame turned a two-body problem into a one-body one. That is what relative velocity is for.

10. Uniform circular motion

Constant speed, continuously changing direction — so the motion is accelerated, which is the trap flagged back in section 4.

$\omega =\frac{d\theta }{dt}=\frac{2\pi }{T}=2\pi f,v=\omega r$

$a_c=\frac{v^2}{r}={\omega }^{2}r$

The acceleration points to the centre, perpendicular to the velocity at every instant. That perpendicularity is exactly why the speed stays fixed: a force along the motion would change speed, and one across it can only change direction.

Trap. $a_c=v^2/r$ says the acceleration rises as the radius shrinks at fixed speed. A tight turn is more punishing than a wide one at the same speed, which is why racing lines are drawn as wide as the track allows.

Illustration 10

A particle moves in a circle of radius $0.5\text{m}$ at 4 revolutions per second. Find its angular speed, linear speed, centripetal acceleration and period.

$\omega =2\pi f=8\pi =25.1\text{rad/s}$

$v=\omega r=25.1(0.5)=12.6\text{m/s}$

$a_c={\omega }^{2}r=(25.1{)}^2(0.5)=316{\text{m/s}}^2$

$T=\frac{1}{f}=0.25\text{s}$

Check by the other route: $a_c=\frac{v^2}{r}=\frac{158}{0.5}=316{\text{m/s}}^2$. The two forms agree.

Scale check: that is about 32g. Modest speeds on a small radius produce enormous accelerations, which is how a centrifuge works.

Summary

• Average speed is weighted by time, never distance. Equal distances $\Rightarrow$ harmonic mean; equal times $\Rightarrow$ arithmetic mean.
• distance $\ge$ |displacement|, equality only when motion never reverses.
• Any resultant lies between $|A-B|$ and $A+B$ — check every vector answer against that.
• Dot product zero means perpendicular; cross product zero means parallel. $|\vec{A}||\vec{B}|\sin \theta$ cross-checks the angle the dot product gave.
• Acceleration comes from a change in magnitude or direction — uniform circular motion is accelerated.
• The four constant-$a$ equations are valid only for constant $a$. Otherwise, integrate.
• Choose one positive direction and never switch it mid-problem.
• $v$–$t$ graph: slope $=a$, signed area $=s$, unsigned area $=$ distance.
• Under gravity: $t_{\text{up}}=u/g$, $H=u^2/2g$, $T=2u/g$, return speed $=u$.
• ${\vec{v}}_{AB}={\vec{v}}_A-{\vec{v}}_B$. Shortest time: aim across. Shortest path: aim upstream.
• Projectile: horizontal and vertical are independent. $R$ peaks at 45°, and $\theta$ and $90°-\theta$ share a range.
• At the peak of a projectile the velocity is horizontal, not zero.
• Horizontal projection: the fall time is set by the height alone, so a ball rolled off a table lands with one simply dropped.
• Uniform circular motion: $v=\omega r$, $a_c=v^2/r={\omega }^{2}r$ toward the centre, perpendicular to $v$ — which is why the speed holds constant.