CDS Physics · Kinematics and Motion

Motion in a Plane

Adding vectors and splitting them into components, a body falling or launched sideways under gravity, and why uniform circular motion is accelerated.

All Kinematics and Motion notes

Why this matters

Eight CDS questions from 2018 (II) to 2026 (II), and all three of the chapter's HARD questions are here: spherical polar components, the velocity and acceleration of a position vector, and the largest and smallest resultant of four forces.

Concept 1 of 3: Adding vectors and resolving them

A vector has a size and a direction, so two vectors cannot simply be added as numbers. Two forces pulling the same way add fully; pulling opposite ways, they cancel as far as they can; at right angles, the resultant is the slanting side of a right triangle. Any vector can also be split into parts along the x, y and z axes, and those parts can be added axis by axis.

Definition

Rules for vectors:

Resultant of $A$ and $B$ at angle $\theta$ : $R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$ .
$R$ is largest, $A + B$ , at $\theta = 0°$ ; smallest, $|A - B|$ , at $\theta = 180°$ ; $\sqrt{A^{2} + B^{2}}$ at 90°.
So for any angle other than 0°, $|\vec{A} + \vec{B}| < |\vec{A}| + |\vec{B}|$ .
Components in a plane: a vector of size $P$ at angle $\phi$ to the x-axis has $P\cos\phi$ along x and $P\sin\phi$ along y.
In spherical polar form (polar angle $\theta$ from the z-axis, azimuth $\alpha$ from the x-axis): $P_x = P\sin\theta\cos\alpha$ , $P_y = P\sin\theta\sin\alpha$ , $P_z = P\cos\theta$ .
Velocity is the rate of change of the position vector and acceleration the rate of change of velocity: differentiate each component separately. A constant component disappears.

Resultant of two vectors

R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta} \qquad |A - B| \le R \le A + B \qquad \vec{v} = \dfrac{d\vec{r}}{dt}, \ \vec{a} = \dfrac{d\vec{v}}{dt}

A, Bsizes of the two vectors
θangle between them
Rsize of the resultant

Worked example

Forces of 5 N and 12 N act on a body. Find the resultant when they act along the same line in the same direction, in opposite directions, and at right angles.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2019 · CDS (I) 2019 — General Knowledge · Q62Moderate

Example 1 · Kinematics and Motion · Motion in a Plane

If two vectors

\vec{A}

and

\vec{B}

are at an angle

\theta \neq 0^\circ

, then

Sizes add only when the directions match

|\vec{A}| + |\vec{B}|

equals

|\vec{A} + \vec{B}|

only at 0°. At any other angle the resultant is smaller than the sum of the sizes, never larger.

Combine forces along one line first

When several forces act along two perpendicular lines, first add the forces on each line (same way: add; opposite ways: subtract). Only then join the two line totals with Pythagoras.

sin and cos swap with the angle you measure from

Measured from the x-axis, the x-part uses cos and the y-part sin. Measured from the z-axis (the polar angle), the z-part uses cos and the part in the xy-plane uses sin.

Drill 3 more on adding vectors and resolving them

Concept 2 of 3: Free fall and horizontal launch

Near the earth every falling body speeds up by the same amount each second: its acceleration g stays the same all the way down. A ball thrown sideways falls exactly like a ball simply dropped, because gravity pulls only downward and does nothing to the sideways motion. So both reach the ground together, and the sideways speed only decides how far away the ball lands.

Definition

Motion under gravity alone (air resistance ignored):

The acceleration stays constant at $g$ (about 9.8 m/s², often taken as 10 m/s²). Speed, momentum and kinetic energy increase; potential energy decreases.
Dropped from height $h$ : $h = \tfrac{1}{2}gt^{2}$ , so $t = \sqrt{2h/g}$ , and it lands at $v = \sqrt{2gh}$ .
Launched horizontally with speed $u$ from height $h$ : the time to land is the same $\sqrt{2h/g}$ , whatever $u$ is.
The horizontal speed stays $u$ throughout, so the body lands at a horizontal distance $R = u\sqrt{2h/g}$ .

Free fall and horizontal launch

h = \tfrac{1}{2}gt^{2} \qquad t = \sqrt{\dfrac{2h}{g}} \qquad v = \sqrt{2gh} \qquad R = u\sqrt{\dfrac{2h}{g}}

hheight of the fall
gacceleration due to gravity
uhorizontal launch speed
Rhorizontal distance to the landing point

Worked example

A ball rolls off a table 20 m high with a horizontal speed of 15 m/s. When and how far from the foot of the table does it land? (Take

g = 10

m/s².)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2026 · CDS (I) 2026 — General Knowledge · Q19Moderate

Example 2 · Kinematics and Motion · Motion in a Plane

An object is launched horizontally from a height of 45 m above the ground with a speed of 10 m/s. The time at which the object will hit the ground is : (Take

g = 10\ \text{m/s}^2

)

The launch speed does not change the time in the air

For a horizontal launch the time comes from the height alone,

t = \sqrt{2h/g}

. The horizontal speed changes where the body lands, not when.

Acceleration stays constant, speed does not

In free fall the acceleration

g

is the one thing that stays uniform. Speed, momentum and kinetic energy all grow as the body falls.

Drill 1 more on free fall and horizontal launch

Concept 3 of 3: Uniform circular motion

A car going round a bend at a steady speed is still accelerating, because its direction keeps changing, and velocity includes direction. The push that keeps turning it always points to the centre of the circle, at right angles to the way the car is moving. Its size stays the same, but its direction swings round with the car, so even the acceleration is not constant.

Definition

For a body moving in a circle of radius $r$ at constant speed $v$ :

The speed is constant but the velocity changes, because its direction changes.
The acceleration is centripetal: it points to the centre, perpendicular to the velocity.
Its size is $a = \dfrac{v^{2}}{r} = \omega^{2}r$ , with $\omega = v/r$ the angular speed.
Its size is constant but its direction changes, so the acceleration is non-zero and not constant.
Time for one round: $T = \dfrac{2\pi r}{v}$ .

Centripetal acceleration

a = \dfrac{v^{2}}{r} = \omega^{2}r \qquad v = \omega r \qquad T = \dfrac{2\pi r}{v}

vspeed along the circle
rradius of the circle
ωangular speed in radians per second
Ttime for one round

Worked example

A car goes round a circular bend of radius 50 m at a steady 20 m/s. Find the size of its acceleration and say where it points.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (I) 2021 — General Knowledge · Q118Easy

Example 3 · Kinematics and Motion · Motion in a Plane

The direction of acceleration in uniform circular motion is along the

Uniform circular motion is accelerated

Constant speed does not mean zero acceleration. The direction of the velocity keeps changing, so there is always an acceleration of size

v^{2}/r

Constant size is not constant acceleration

The centripetal acceleration has a fixed size but its direction turns with the body. A vector whose direction changes is not constant.

Drill 1 more on uniform circular motion

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Adding vectors and resolving them
Resultant of two vectors
$R = \sqrt{A^{2} + B^{2} + 2AB\cos\theta} \qquad |A - B| \le R \le A + B \qquad \vec{v} = \dfrac{d\vec{r}}{dt}, \ \vec{a} = \dfrac{d\vec{v}}{dt}$
Free fall and horizontal launch
Free fall and horizontal launch
$h = \tfrac{1}{2}gt^{2} \qquad t = \sqrt{\dfrac{2h}{g}} \qquad v = \sqrt{2gh} \qquad R = u\sqrt{\dfrac{2h}{g}}$
Uniform circular motion
Centripetal acceleration
$a = \dfrac{v^{2}}{r} = \omega^{2}r \qquad v = \omega r \qquad T = \dfrac{2\pi r}{v}$

Watch out for (7)

Sizes add only when the directions match
→ Adding vectors and resolving them
Combine forces along one line first
→ Adding vectors and resolving them
sin and cos swap with the angle you measure from
→ Adding vectors and resolving them
The launch speed does not change the time in the air
→ Free fall and horizontal launch
Acceleration stays constant, speed does not
→ Free fall and horizontal launch
Uniform circular motion is accelerated
→ Uniform circular motion
Constant size is not constant acceleration
→ Uniform circular motion

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