Module 1: Kinematics

Describe and predict motion using displacement, velocity and acceleration, master motion graphs and the equations of motion, and build the vector toolkit for two-dimensional motion.

Pillar 1 of 2

Motion in a Straight Line

Kinematics is the language of motion. Before asking why things move (that is dynamics), we learn to describe motion precisely using displacement, velocity and acceleration, and to read and build the graphs and equations that predict where an object will be and how fast it is going. Everything in this pillar is one-dimensional, so direction is captured by a single $+$ or $\;-$ sign.

Distance and Displacement

Position is where an object sits relative to a chosen reference point (the origin). As an object moves, two different quantities describe how far it has gone, and confusing them is one of the most common early mistakes in kinematics.

Distance (scalar)

The total length of the path travelled. It has magnitude only, is always positive, and depends on the whole route taken.

Displacement (vector)

The change in position, measured in a straight line from start to finish. It has a direction, so in one dimension we give it a $+$ or $-$ sign.

Walk 5 m east then 2 m back west: the distance travelled is 7 m, but the displacement (start to end) is only 3 m east.

Displacement

\Delta s = s_{\text{final}} - s_{\text{initial}}

Worked Example

A delivery robot moves $6\text{ m}$ east, then $10\text{ m}$ west along a corridor. Find the total distance travelled and its displacement. (Take east as positive.)

Distance (add path lengths):

$d = 6 + 10 = 16\text{ m}$

Displacement (add signed vectors):

$\Delta s = (+6) + (-10) = -4\text{ m}$

$\Delta s = \mathbf{4\text{ m west}}$

The negative sign is not "less than zero distance", it simply means the final position is west of the start.

Practice Question

Q: An athlete runs exactly one lap of a $400\text{ m}$ track, finishing where they started. What distance did they cover, and what was their displacement?

Show Answer

Distance = whole path run = $400\text{ m}$ .

Displacement = straight line from start to finish = $0\text{ m}$ , because they end exactly where they began.

A large distance can have zero displacement, which is why the two must never be confused.

Speed and Velocity

Speed is how fast distance is covered (a scalar). Velocity is how fast displacement changes, and in which direction (a vector). Average values use the total change over the total time, while the instantaneous value is the value at a single moment, which is what a car's speedometer shows.

Average

The overall rate across an entire trip: total change divided by total time. It smooths over any speeding up and slowing down in between.

Instantaneous

The value at one instant, found from the gradient of a displacement-time graph at that exact point.

Average Speed and Velocity

Average speed

v = \frac{d}{t}

Average velocity

v_{\text{av}} = \frac{\Delta s}{\Delta t}

Worked Example

A cyclist rides $300\text{ m}$ north in $20\text{ s}$ , then $\;100\text{ m}$ south in $10\text{ s}$ . Find the average speed and the average velocity for the whole trip.

Total distance and displacement:

$d = 300 + 100 = 400\text{ m}$

$\Delta s = 300 - 100 = 200\text{ m north}$

Solution:

$\text{speed} = \frac{400}{30} = \mathbf{13.3\text{ m/s}}$

$v_{\text{av}} = \frac{200}{30} = \mathbf{6.7\text{ m/s north}}$

The average velocity is smaller because the return leg cancels part of the displacement.

Practice Question

Q: A train travels $240\text{ km}$ east in $3\text{ h}$ , then $\;120\text{ km}$ west in $2\text{ h}$ . Find its average speed and average velocity in $\text{km/h}$ .

Show Answer

Total distance $= 240 + 120 = 360\text{ km}$ over $5\text{ h}$ :

$\text{speed} = \frac{360}{5} = 72\text{ km/h}$

Displacement $= 240 - 120 = 120\text{ km east}$ :

$v_{\text{av}} = \frac{120}{5} = 24\text{ km/h east}$

Acceleration

Acceleration is the rate of change of velocity, and it is a vector. An object accelerates whenever its speed changes, its direction changes, or both. A negative acceleration in the direction of motion means the object is slowing down (decelerating).

Acceleration

a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}

$u$ = initial velocity, $v$ = final velocity
$a$ is measured in $\text{m/s}^2$

Same sign as velocity

If $a$ points the same way as the motion, the object speeds up.

Opposite sign to velocity

If $a$ opposes the motion, the object slows down. This is what we loosely call deceleration.

Worked Example

A car speeds up from $10\text{ m/s}$ to $30\text{ m/s}$ in $\;5\text{ s}$ . Find its acceleration.

Solution:

$a = \frac{v - u}{t} = \frac{30 - 10}{5}$

$a = \mathbf{4\text{ m/s}^2}$

Practice Question

Q: A train travelling at $25\text{ m/s}$ slows to $5\text{ m/s}$ in $\;4\text{ s}$ . Find its acceleration and explain the sign.

Show Answer

$a = \frac{v - u}{t} = \frac{5 - 25}{4} = -5\text{ m/s}^2$

The acceleration is negative because it points opposite to the motion. The train is decelerating at $5\text{ m/s}^2$ .

Displacement-Time Graphs

On a displacement-time graph, the gradient equals the velocity. Reading the shape of the line tells you exactly how the object is moving: a flat line is at rest, a straight slope is constant velocity, and a curve means the velocity is changing (acceleration).

On a displacement–time graph the gradient is the velocity; a steeper line means a faster object, and a curve means the velocity is changing.

Velocity is the Gradient

v = \text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta s}{\Delta t}

To find the instantaneous velocity at a point on a curve, draw a tangent to the curve at that point and measure its gradient. A steeper tangent means a faster instantaneous velocity.

Worked Example

A drone's displacement-time graph is a straight line from $(0\text{ s}, 0\text{ m})$ to $\;(4\text{ s}, 20\text{ m})$ , and then stays flat at $20\text{ m}$ until $6\text{ s}$ . Find the velocity in each phase.

Phase 1 (sloped line):

$v = \frac{\Delta s}{\Delta t} = \frac{20 - 0}{4 - 0} = \mathbf{5\text{ m/s}}$

Phase 2 (flat line):

$v = \frac{0}{2} = \mathbf{0\text{ m/s (at rest)}}$

Practice Question

Q: On a displacement-time graph, an object's line curves upward and gets steeper and steeper. Describe how the object is moving.

Show Answer

The gradient is the velocity. A gradient that grows steeper means the velocity is increasing, so the object is accelerating (speeding up) in the positive direction.

Velocity-Time Graphs

A velocity-time graph is doubly useful: the gradient equals the acceleration, and the area under the graph equals the displacement. Split the area into rectangles and triangles and add them up (area below the time axis counts as negative displacement).

Two Readings from One Graph

Gradient = acceleration

a = \frac{\Delta v}{\Delta t}

Area = displacement

s = \text{area under graph}

Worked Example

An object moves at a constant $8\text{ m/s}$ for $5\text{ s}$ , then decelerates uniformly to rest over the next $4\text{ s}$ . Find the total displacement from the area under its velocity-time graph.

The shaded area is the displacement: a 40 m rectangle plus a 16 m triangle.

Solution (rectangle + triangle):

$\text{rectangle} = 8 \times 5 = 40\text{ m}$

$\text{triangle} = \tfrac{1}{2} \times 4 \times 8 = 16\text{ m}$

$s = 40 + 16 = \mathbf{56\text{ m}}$

Practice Question

Q: A cyclist accelerates uniformly from rest to $12\text{ m/s}$ in $\;3\text{ s}$ , then holds $12\text{ m/s}$ for $\;2\text{ s}$ . Find the acceleration in the first phase and the total displacement.

Show Answer

Acceleration = gradient of phase 1:

$a = \frac{12 - 0}{3} = 4\text{ m/s}^2$

Displacement = triangle + rectangle:

$s = \tfrac{1}{2}(3)(12) + (12)(2) = 18 + 24 = 42\text{ m}$

The Equations of Motion (SUVAT)

When the acceleration is constant, five quantities are linked: displacement $\;s$ , initial velocity $u$ , final velocity $v$ , acceleration $a$ , and time $t$ . Four equations connect them, and each one leaves out a different variable, so you pick the equation that has the three quantities you know plus the one you want.

The Four Equations of Uniform Acceleration

v = u + at

s = ut + \tfrac{1}{2}at^2

v^2 = u^2 + 2as

s = \tfrac{1}{2}(u + v)t

Where the first two come from

The first equation is just the definition of acceleration rearranged, and the fourth is displacement as average velocity times time:

a = \frac{v - u}{t} \;\Rightarrow\; v = u + at

s = \left(\frac{u + v}{2}\right)t

Which equation?

$v = u + at$ — no $s$
$s = ut + \tfrac{1}{2}at^2$ — no $v$
$v^2 = u^2 + 2as$ — no $t$
$s = \tfrac{1}{2}(u+v)t$ — no $a$

Method

1. List $s, u, v, a, t$ and mark the unknown.
2. Fix a sign convention for direction.
3. Pick the equation missing the variable you neither know nor want.

Worked Example

A car starts from rest and accelerates at $3\text{ m/s}^2$ for $8\text{ s}$ . Find its final velocity and the distance it travels.

Given:

$u = 0,\; a = 3\text{ m/s}^2,\; t = 8\text{ s}$

Solution:

$v = u + at = 0 + (3)(8) = \mathbf{24\text{ m/s}}$

$s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(3)(8)^2 = \mathbf{96\text{ m}}$

Worked Example

A motorbike accelerates from $8\text{ m/s}$ to $20\text{ m/s}$ over a distance of $56\text{ m}$ . Find its acceleration. (Time is unknown, so choose the equation without $t$ .)

Given:

$u = 8,\; v = 20,\; s = 56$

Solution:

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

$20^2 = 8^2 + 2a(56)$

$400 - 64 = 112\,a$

$a = \frac{336}{112} = \mathbf{3\text{ m/s}^2}$

Practice Question

Q: A car travelling at $20\text{ m/s}$ brakes at $5\text{ m/s}^2$ . How far does it travel before stopping?

Show Answer

Use the equation without $t$ , with $v = 0$ and $a = -5\text{ m/s}^2$ :

$v^2 = u^2 + 2as \;\Rightarrow\; 0 = 20^2 + 2(-5)s$

$s = \frac{400}{10} = 40\text{ m}$

Free Fall and Acceleration due to Gravity

Near Earth's surface, any object in free fall (with air resistance ignored) accelerates downward at $g \approx 9.8\text{ m/s}^2$ , no matter what its mass is. This means free fall is just uniform-acceleration motion, so the SUVAT equations apply with $a = g$ . Choose a sign convention first; for example, taking up as positive makes $a = -9.8\text{ m/s}^2$ .

Mass does not matter

A feather and a hammer fall at the same rate in a vacuum. Only air resistance makes light objects fall more slowly in real life.

Pick a sign convention

Decide up or down as positive before you start. Keep $u$ , $v$ , $\;s$ and $g$ consistent with that choice.

SUVAT with $a = g$

v = u + gt

s = ut + \tfrac{1}{2}gt^2

v^2 = u^2 + 2gs

Worked Example

A ball is dropped from rest and falls for $3.0\text{ s}$ . Find its velocity just before landing and the height it fell. (Take down as positive, $g = 9.8\text{ m/s}^2$ .)

The ball speeds up uniformly as it falls, so each 1 s drop is longer than the one before.

Solution:

$v = u + gt = 0 + (9.8)(3.0) = \mathbf{29.4\text{ m/s}}$

$s = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = \mathbf{44.1\text{ m}}$

Practice Question

Q: A ball is thrown straight up at $19.6\text{ m/s}$ . How long does it take to reach its highest point?

Show Answer

At the highest point the velocity is momentarily zero. Taking up as positive, $a = -9.8\text{ m/s}^2$ :

$v = u + at \;\Rightarrow\; 0 = 19.6 + (-9.8)t$

$t = \frac{19.6}{9.8} = 2.0\text{ s}$

Pillar 2 of 2

Motion on a Plane

Real motion is rarely a perfectly straight line. To handle motion in two dimensions we treat quantities as vectors, learn to add them, and split them into perpendicular components. This vector toolkit underpins projectile motion, forces and fields in every module that follows.

Scalars and Vectors

A scalar has magnitude only. A vector has magnitude and direction. Vectors are drawn as arrows: the length shows the magnitude and the arrowhead shows the direction. A vector is written in bold or with an arrow, such as $\vec{v}$ .

Scalars

Distance, speed, time, mass, energy, temperature.

Vectors

Displacement, velocity, acceleration, force, momentum.

A vector is drawn as an arrow: its length represents the magnitude, and the angle θ from a reference direction gives its direction.

Worked Example

Classify each quantity as a scalar or a vector, and give the reason: mass $= 5\text{ kg}$ ; velocity $= 3\text{ m/s east}$ ; temperature $= 20^\circ\text{C}$ ; force $= 10\text{ N up}$ .

Mass — scalar (magnitude only).
Velocity — vector (has a direction: east).
Temperature — scalar (no direction).
Force — vector (has a direction: up).

Practice Question

Q: Speed and velocity both describe how fast something moves. Which one is the vector, and why?

Show Answer

Velocity is the vector, because it states a direction as well as a magnitude. Speed is the scalar — it is just the magnitude, with no direction attached.

Adding Vectors Graphically

To add vectors, draw them head-to-tail: the tail of the second starts where the head of the first ends. The resultant is the single vector drawn from the very first tail to the very last head. The order does not change the result. When the two vectors are perpendicular, the resultant comes straight from Pythagoras and trigonometry.

Resultant of Two Perpendicular Vectors

R = \sqrt{A^2 + B^2}

\theta = \tan^{-1}\!\left(\frac{B}{A}\right)

Worked Example

A person walks $3\text{ m}$ east, then $4\text{ m}$ north. Find the magnitude and direction of the resultant displacement.

Head-to-tail: the 3 m east and 4 m north legs give a 5 m resultant at 53° north of east.

Solution:

$R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ m}$

$\theta = \tan^{-1}\!\left(\frac{4}{3}\right) = 53^\circ\text{ north of east}$

Practice Question

Q: A swimmer swims $5\text{ m}$ east, then $12\text{ m}$ north. Find the magnitude and direction of the resultant displacement.

Show Answer

$R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ m}$

$\theta = \tan^{-1}\!\left(\frac{12}{5}\right) = 67^\circ\text{ north of east}$

Resolving Vectors into Components

Any vector can be split into two perpendicular components, usually a horizontal ( $x$ ) and a vertical ( $y$ ) part. For a vector of magnitude $\;A$ at angle $\theta$ to the horizontal, use the sides of the right-angled triangle it forms.

Components

A_x = A\cos\theta

A_y = A\sin\theta

Worked Example

A force of $50\text{ N}$ acts at $30^\circ$ above the horizontal. Find its horizontal and vertical components.

The 50 N force resolves into a horizontal part Fₓ = 50 cos 30° and a vertical part F_y = 50 sin 30°.

Solution:

$F_x = 50\cos 30^\circ = 43.3\text{ N}$

$F_y = 50\sin 30^\circ = 25\text{ N}$

Practice Question

Q: A $100\text{ N}$ force acts at $60^\circ$ above the horizontal. Find its horizontal and vertical components.

Show Answer

$F_x = 100\cos 60^\circ = 50\text{ N}$

$F_y = 100\sin 60^\circ = 86.6\text{ N}$

Adding Vectors by Components

For an exact answer, or when vectors are not perpendicular, use the component method:

Resolve every vector into its $x$ and $y$ components.
Add all the $x$ components, and separately add all the $y$ components.
Recombine the totals with Pythagoras and $\tan^{-1}$ to get the resultant.

Recombining the Totals

R = \sqrt{(\Sigma x)^2 + (\Sigma y)^2}

\theta = \tan^{-1}\!\left(\frac{\Sigma y}{\Sigma x}\right)

Worked Example

A hiker walks $20\text{ m}$ at $30^\circ$ north of east, then $\;20\text{ m}$ at $30^\circ$ north of west. Find the resultant displacement.

Resolve (east and north positive):

$x\text{: } 20\cos 30^\circ - 20\cos 30^\circ = 0$

$y\text{: } 20\sin 30^\circ + 20\sin 30^\circ = 10 + 10 = 20\text{ m}$

Recombine:

$R = \sqrt{0^2 + 20^2} = 20\text{ m due north}$

The eastward and westward parts cancel, leaving a purely northward resultant.

Practice Question

Q: Two forces act at a point: $5\text{ N}$ east and $5\text{ N}$ north. Use the component method to find the resultant force.

Show Answer

$\Sigma x = 5\text{ N}, \; \Sigma y = 5\text{ N}$

$R = \sqrt{5^2 + 5^2} = \sqrt{50} = 7.1\text{ N}$

$\theta = \tan^{-1}\!\left(\frac{5}{5}\right) = 45^\circ\text{ north of east}$

Relative Velocity

Every velocity is measured relative to a frame of reference. The velocity of object A as seen fromobject B is found by vector subtraction. In one dimension you subtract using signs; in two dimensions you subtract the components.

Relative Velocity

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

Worked Example

A boat heads due north across a river at $4\text{ m/s}$ relative to the water, while the river flows east at $3\text{ m/s}$ . Find the boat's velocity relative to the ground.

Adding the boat's 4 m/s north and the current's 3 m/s east gives 5 m/s at 37° east of north.

Solution (the two velocities are perpendicular):

$v = \sqrt{4^2 + 3^2} = 5\text{ m/s}$

$\theta = \tan^{-1}\!\left(\frac{3}{4}\right) = 37^\circ\text{ east of north}$

The current carries the boat downstream, so it crosses at an angle rather than straight across.

Practice Question

Q: Car A drives east at $20\text{ m/s}$ and car B drives west at $\;15\text{ m/s}$ directly toward it. What is the velocity of A relative to B?

Show Answer

Take east as positive, so $v_A = +20$ and $v_B = -15\text{ m/s}$ :

$v_{AB} = v_A - v_B = 20 - (-15) = 35\text{ m/s east}$

They close on each other at $35\text{ m/s}$ , which is why head-on approaches feel so fast.

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Pillar 1 of 2

Motion in a Straight Line

Distance and Displacement

Distance (scalar)

The total length of the path travelled. It has magnitude only, is always positive, and depends on the whole route taken.

Displacement (vector)

The change in position, measured in a straight line from start to finish. It has a direction, so in one dimension we give it a $+$ or $-$ sign.

Walk 5 m east then 2 m back west: the distance travelled is 7 m, but the displacement (start to end) is only 3 m east.

Displacement

\Delta s = s_{\text{final}} - s_{\text{initial}}

Worked Example

A delivery robot moves $6\text{ m}$ east, then $10\text{ m}$ west along a corridor. Find the total distance travelled and its displacement. (Take east as positive.)

Distance (add path lengths):

$d = 6 + 10 = 16\text{ m}$

Displacement (add signed vectors):

$\Delta s = (+6) + (-10) = -4\text{ m}$

$\Delta s = \mathbf{4\text{ m west}}$

The negative sign is not "less than zero distance", it simply means the final position is west of the start.

Practice Question

Q: An athlete runs exactly one lap of a $400\text{ m}$ track, finishing where they started. What distance did they cover, and what was their displacement?

Show Answer

Distance = whole path run = $400\text{ m}$ .

Displacement = straight line from start to finish = $0\text{ m}$ , because they end exactly where they began.

A large distance can have zero displacement, which is why the two must never be confused.

Speed and Velocity

Average

The overall rate across an entire trip: total change divided by total time. It smooths over any speeding up and slowing down in between.

Instantaneous

The value at one instant, found from the gradient of a displacement-time graph at that exact point.

Average Speed and Velocity

Average speed

v = \frac{d}{t}

Average velocity

v_{\text{av}} = \frac{\Delta s}{\Delta t}

Worked Example

A cyclist rides $300\text{ m}$ north in $20\text{ s}$ , then $\;100\text{ m}$ south in $10\text{ s}$ . Find the average speed and the average velocity for the whole trip.

Total distance and displacement:

$d = 300 + 100 = 400\text{ m}$

$\Delta s = 300 - 100 = 200\text{ m north}$

Solution:

$\text{speed} = \frac{400}{30} = \mathbf{13.3\text{ m/s}}$

$v_{\text{av}} = \frac{200}{30} = \mathbf{6.7\text{ m/s north}}$

The average velocity is smaller because the return leg cancels part of the displacement.

Practice Question

Q: A train travels $240\text{ km}$ east in $3\text{ h}$ , then $\;120\text{ km}$ west in $2\text{ h}$ . Find its average speed and average velocity in $\text{km/h}$ .

Show Answer

Total distance $= 240 + 120 = 360\text{ km}$ over $5\text{ h}$ :

$\text{speed} = \frac{360}{5} = 72\text{ km/h}$

Displacement $= 240 - 120 = 120\text{ km east}$ :

$v_{\text{av}} = \frac{120}{5} = 24\text{ km/h east}$

Acceleration

a = \frac{\Delta v}{\Delta t} = \frac{v - u}{t}

$u$ = initial velocity, $v$ = final velocity
$a$ is measured in $\text{m/s}^2$

Same sign as velocity

If $a$ points the same way as the motion, the object speeds up.

Opposite sign to velocity

If $a$ opposes the motion, the object slows down. This is what we loosely call deceleration.

Worked Example

A car speeds up from $10\text{ m/s}$ to $30\text{ m/s}$ in $\;5\text{ s}$ . Find its acceleration.

Solution:

$a = \frac{v - u}{t} = \frac{30 - 10}{5}$

$a = \mathbf{4\text{ m/s}^2}$

Practice Question

Q: A train travelling at $25\text{ m/s}$ slows to $5\text{ m/s}$ in $\;4\text{ s}$ . Find its acceleration and explain the sign.

Show Answer

$a = \frac{v - u}{t} = \frac{5 - 25}{4} = -5\text{ m/s}^2$

The acceleration is negative because it points opposite to the motion. The train is decelerating at $5\text{ m/s}^2$ .

Displacement-Time Graphs

On a displacement–time graph the gradient is the velocity; a steeper line means a faster object, and a curve means the velocity is changing.

Velocity is the Gradient

v = \text{gradient} = \frac{\text{rise}}{\text{run}} = \frac{\Delta s}{\Delta t}

To find the instantaneous velocity at a point on a curve, draw a tangent to the curve at that point and measure its gradient. A steeper tangent means a faster instantaneous velocity.

Worked Example

Phase 1 (sloped line):

$v = \frac{\Delta s}{\Delta t} = \frac{20 - 0}{4 - 0} = \mathbf{5\text{ m/s}}$

Phase 2 (flat line):

$v = \frac{0}{2} = \mathbf{0\text{ m/s (at rest)}}$

Practice Question

Q: On a displacement-time graph, an object's line curves upward and gets steeper and steeper. Describe how the object is moving.

Show Answer

The gradient is the velocity. A gradient that grows steeper means the velocity is increasing, so the object is accelerating (speeding up) in the positive direction.

Velocity-Time Graphs

Two Readings from One Graph

Gradient = acceleration

a = \frac{\Delta v}{\Delta t}

Area = displacement

s = \text{area under graph}

Worked Example

The shaded area is the displacement: a 40 m rectangle plus a 16 m triangle.

Solution (rectangle + triangle):

$\text{rectangle} = 8 \times 5 = 40\text{ m}$

$\text{triangle} = \tfrac{1}{2} \times 4 \times 8 = 16\text{ m}$

$s = 40 + 16 = \mathbf{56\text{ m}}$

Practice Question

Show Answer

Acceleration = gradient of phase 1:

$a = \frac{12 - 0}{3} = 4\text{ m/s}^2$

Displacement = triangle + rectangle:

$s = \tfrac{1}{2}(3)(12) + (12)(2) = 18 + 24 = 42\text{ m}$

The Equations of Motion (SUVAT)

The Four Equations of Uniform Acceleration

v = u + at

s = ut + \tfrac{1}{2}at^2

v^2 = u^2 + 2as

s = \tfrac{1}{2}(u + v)t

Where the first two come from

The first equation is just the definition of acceleration rearranged, and the fourth is displacement as average velocity times time:

a = \frac{v - u}{t} \;\Rightarrow\; v = u + at

s = \left(\frac{u + v}{2}\right)t

Which equation?

$v = u + at$ — no $s$
$s = ut + \tfrac{1}{2}at^2$ — no $v$
$v^2 = u^2 + 2as$ — no $t$
$s = \tfrac{1}{2}(u+v)t$ — no $a$

Method

1. List $s, u, v, a, t$ and mark the unknown.
2. Fix a sign convention for direction.
3. Pick the equation missing the variable you neither know nor want.

Worked Example

A car starts from rest and accelerates at $3\text{ m/s}^2$ for $8\text{ s}$ . Find its final velocity and the distance it travels.

Given:

$u = 0,\; a = 3\text{ m/s}^2,\; t = 8\text{ s}$

Solution:

$v = u + at = 0 + (3)(8) = \mathbf{24\text{ m/s}}$

$s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2}(3)(8)^2 = \mathbf{96\text{ m}}$

Worked Example

A motorbike accelerates from $8\text{ m/s}$ to $20\text{ m/s}$ over a distance of $56\text{ m}$ . Find its acceleration. (Time is unknown, so choose the equation without $t$ .)

Given:

$u = 8,\; v = 20,\; s = 56$

Solution:

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

$20^2 = 8^2 + 2a(56)$

$400 - 64 = 112\,a$

$a = \frac{336}{112} = \mathbf{3\text{ m/s}^2}$

Practice Question

Q: A car travelling at $20\text{ m/s}$ brakes at $5\text{ m/s}^2$ . How far does it travel before stopping?

Show Answer

Use the equation without $t$ , with $v = 0$ and $a = -5\text{ m/s}^2$ :

$v^2 = u^2 + 2as \;\Rightarrow\; 0 = 20^2 + 2(-5)s$

$s = \frac{400}{10} = 40\text{ m}$

Free Fall and Acceleration due to Gravity

Mass does not matter

A feather and a hammer fall at the same rate in a vacuum. Only air resistance makes light objects fall more slowly in real life.

Pick a sign convention

Decide up or down as positive before you start. Keep $u$ , $v$ , $\;s$ and $g$ consistent with that choice.

SUVAT with $a = g$

v = u + gt

s = ut + \tfrac{1}{2}gt^2

v^2 = u^2 + 2gs

Worked Example

A ball is dropped from rest and falls for $3.0\text{ s}$ . Find its velocity just before landing and the height it fell. (Take down as positive, $g = 9.8\text{ m/s}^2$ .)

The ball speeds up uniformly as it falls, so each 1 s drop is longer than the one before.

Solution:

$v = u + gt = 0 + (9.8)(3.0) = \mathbf{29.4\text{ m/s}}$

$s = ut + \tfrac{1}{2}gt^2 = 0 + \tfrac{1}{2}(9.8)(3.0)^2 = \mathbf{44.1\text{ m}}$

Practice Question

Q: A ball is thrown straight up at $19.6\text{ m/s}$ . How long does it take to reach its highest point?

Show Answer

At the highest point the velocity is momentarily zero. Taking up as positive, $a = -9.8\text{ m/s}^2$ :

$v = u + at \;\Rightarrow\; 0 = 19.6 + (-9.8)t$

$t = \frac{19.6}{9.8} = 2.0\text{ s}$

Pillar 2 of 2

Motion on a Plane

Scalars and Vectors

Scalars

Distance, speed, time, mass, energy, temperature.

Vectors

Displacement, velocity, acceleration, force, momentum.

A vector is drawn as an arrow: its length represents the magnitude, and the angle θ from a reference direction gives its direction.

Worked Example

Classify each quantity as a scalar or a vector, and give the reason: mass $= 5\text{ kg}$ ; velocity $= 3\text{ m/s east}$ ; temperature $= 20^\circ\text{C}$ ; force $= 10\text{ N up}$ .

Mass — scalar (magnitude only).
Velocity — vector (has a direction: east).
Temperature — scalar (no direction).
Force — vector (has a direction: up).

Practice Question

Q: Speed and velocity both describe how fast something moves. Which one is the vector, and why?

Show Answer

Velocity is the vector, because it states a direction as well as a magnitude. Speed is the scalar — it is just the magnitude, with no direction attached.

Adding Vectors Graphically

Resultant of Two Perpendicular Vectors

R = \sqrt{A^2 + B^2}

\theta = \tan^{-1}\!\left(\frac{B}{A}\right)

Worked Example

A person walks $3\text{ m}$ east, then $4\text{ m}$ north. Find the magnitude and direction of the resultant displacement.

Head-to-tail: the 3 m east and 4 m north legs give a 5 m resultant at 53° north of east.

Solution:

$R = \sqrt{3^2 + 4^2} = \sqrt{25} = 5\text{ m}$

$\theta = \tan^{-1}\!\left(\frac{4}{3}\right) = 53^\circ\text{ north of east}$

Practice Question

Q: A swimmer swims $5\text{ m}$ east, then $12\text{ m}$ north. Find the magnitude and direction of the resultant displacement.

Show Answer

$R = \sqrt{5^2 + 12^2} = \sqrt{169} = 13\text{ m}$

$\theta = \tan^{-1}\!\left(\frac{12}{5}\right) = 67^\circ\text{ north of east}$

Resolving Vectors into Components

Components

A_x = A\cos\theta

A_y = A\sin\theta

Worked Example

A force of $50\text{ N}$ acts at $30^\circ$ above the horizontal. Find its horizontal and vertical components.

The 50 N force resolves into a horizontal part Fₓ = 50 cos 30° and a vertical part F_y = 50 sin 30°.

Solution:

$F_x = 50\cos 30^\circ = 43.3\text{ N}$

$F_y = 50\sin 30^\circ = 25\text{ N}$

Practice Question

Q: A $100\text{ N}$ force acts at $60^\circ$ above the horizontal. Find its horizontal and vertical components.

Show Answer

$F_x = 100\cos 60^\circ = 50\text{ N}$

$F_y = 100\sin 60^\circ = 86.6\text{ N}$

Adding Vectors by Components

For an exact answer, or when vectors are not perpendicular, use the component method:

Resolve every vector into its $x$ and $y$ components.
Add all the $x$ components, and separately add all the $y$ components.
Recombine the totals with Pythagoras and $\tan^{-1}$ to get the resultant.

Recombining the Totals

R = \sqrt{(\Sigma x)^2 + (\Sigma y)^2}

\theta = \tan^{-1}\!\left(\frac{\Sigma y}{\Sigma x}\right)

Worked Example

A hiker walks $20\text{ m}$ at $30^\circ$ north of east, then $\;20\text{ m}$ at $30^\circ$ north of west. Find the resultant displacement.

Resolve (east and north positive):

$x\text{: } 20\cos 30^\circ - 20\cos 30^\circ = 0$

$y\text{: } 20\sin 30^\circ + 20\sin 30^\circ = 10 + 10 = 20\text{ m}$

Recombine:

$R = \sqrt{0^2 + 20^2} = 20\text{ m due north}$

The eastward and westward parts cancel, leaving a purely northward resultant.

Practice Question

Q: Two forces act at a point: $5\text{ N}$ east and $5\text{ N}$ north. Use the component method to find the resultant force.

Show Answer

$\Sigma x = 5\text{ N}, \; \Sigma y = 5\text{ N}$

$R = \sqrt{5^2 + 5^2} = \sqrt{50} = 7.1\text{ N}$

$\theta = \tan^{-1}\!\left(\frac{5}{5}\right) = 45^\circ\text{ north of east}$

Relative Velocity

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

Worked Example

A boat heads due north across a river at $4\text{ m/s}$ relative to the water, while the river flows east at $3\text{ m/s}$ . Find the boat's velocity relative to the ground.

Adding the boat's 4 m/s north and the current's 3 m/s east gives 5 m/s at 37° east of north.

Solution (the two velocities are perpendicular):

$v = \sqrt{4^2 + 3^2} = 5\text{ m/s}$

$\theta = \tan^{-1}\!\left(\frac{3}{4}\right) = 37^\circ\text{ east of north}$

The current carries the boat downstream, so it crosses at an angle rather than straight across.

Practice Question

Q: Car A drives east at $20\text{ m/s}$ and car B drives west at $\;15\text{ m/s}$ directly toward it. What is the velocity of A relative to B?

Show Answer

Take east as positive, so $v_A = +20$ and $v_B = -15\text{ m/s}$ :

$v_{AB} = v_A - v_B = 20 - (-15) = 35\text{ m/s east}$

They close on each other at $35\text{ m/s}$ , which is why head-on approaches feel so fast.

Practice test

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25 exam-style questions, ordered easy to hard, with instant feedback and full worked solutions. About 50 minutes.

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Motion in a Straight Line

Distance and Displacement

Distance (scalar)

Displacement (vector)

Displacement

Worked Example

Practice Question

Speed and Velocity

Average

Instantaneous

Average Speed and Velocity

Worked Example

Practice Question

Acceleration

Acceleration

Same sign as velocity

Opposite sign to velocity

Worked Example

Practice Question

Displacement-Time Graphs

Velocity is the Gradient

Worked Example

Practice Question

Velocity-Time Graphs

Two Readings from One Graph

Worked Example

Practice Question

The Equations of Motion (SUVAT)

The Four Equations of Uniform Acceleration

Where the first two come from

Which equation?

Method

Worked Example

Worked Example

Practice Question

Free Fall and Acceleration due to Gravity

Mass does not matter

Pick a sign convention

SUVAT with a=ga = ga=g

Worked Example

Practice Question

Motion on a Plane

Scalars and Vectors

Scalars

Vectors

Worked Example

Practice Question

Adding Vectors Graphically

Resultant of Two Perpendicular Vectors

Worked Example

Practice Question

Resolving Vectors into Components

Components

Worked Example

Practice Question

Adding Vectors by Components

Recombining the Totals

Worked Example

Practice Question

Relative Velocity

Relative Velocity

Worked Example

Practice Question

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Motion in a Straight Line

Distance and Displacement

Distance (scalar)

Displacement (vector)

Displacement

Worked Example

Practice Question

Speed and Velocity

Average

Instantaneous

Average Speed and Velocity

Worked Example

Practice Question

Acceleration

Acceleration

SUVAT with $a = g$

SUVAT with $a = g$