Accelerate your learning

Speed is the distance an object travels every second. It is a scalar quantity because it only contains a magnitude, with no associated direction. The average speed of an object over a journey is given by

$\text{average speed} = \dfrac{\text{total distance}}{\text{time taken}}$

The SI unit of speed is metres per second ($\mathrm{m\,s^{-1}}$), although other units such as $\mathrm{km\,h^{-1}}$ or mph are often used when more convenient.

Velocity is the rate of change of displacement. Because displacement is a vector, velocity is also a vector quantity — it describes both a magnitude and a direction. This means velocity can take negative values: for example, if a ball is thrown upwards at $3\,\mathrm{m\,s^{-1}}$ and (taking upwards as positive) comes back down at $-5\,\mathrm{m\,s^{-1}}$, its speeds at those two moments are still $3\,\mathrm{m\,s^{-1}}$ and $5\,\mathrm{m\,s^{-1}}$ respectively — the sign is only meaningful for velocity, not speed. Two objects can have the same speed but different velocities if they are moving in different directions.

The average velocity $\bar v$ of an object over a time interval is

$\bar v = \dfrac{\Delta x}{\Delta t}$

where $\Delta x$ is the total displacement (change in position, in m) and $\Delta t$ is the total time taken (in s). If the initial velocity $u$ and final velocity $v$ are both known, the average velocity can equivalently be found from

$\bar v = \dfrac{u+v}{2}$

On a displacement–time graph, the average velocity over an interval is the total displacement (read on the $y$-axis) divided by the total time taken (read on the $x$-axis); this works whether the graph is a straight or a curved line.

The instantaneous speed (or velocity) is the speed (or velocity) of an object at one specific moment in time, whether the object is moving at constant velocity or accelerating. On a displacement–time graph, a straight line indicates constant velocity, while a curved line indicates an accelerating (changing-velocity) object. To find the instantaneous velocity at a given point on a displacement–time graph: draw a tangent to the curve at that point, then calculate the gradient of the tangent — that gradient equals the instantaneous velocity at that instant.

Acceleration is the rate of change of velocity. It is a vector quantity, measured in metres per second squared ($\mathrm{m\,s^{-2}}$), and it describes how much an object's velocity changes every second.

The average acceleration of an object is calculated using

$\text{average acceleration} = \dfrac{\text{change in velocity}}{\text{time taken}} \qquad\Rightarrow\qquad a = \dfrac{\Delta v}{\Delta t}$

where $a$ is the average acceleration ($\mathrm{m\,s^{-2}}$), $\Delta v$ is the change in velocity ($\mathrm{m\,s^{-1}}$), and $\Delta t$ is the total time taken (s). The change in velocity is the difference between the final and initial velocity:

$\Delta v = v - u$

The instantaneous acceleration is the acceleration of an object at one specific point in time, which is useful for an object whose acceleration is itself constantly changing. On a velocity–time graph, a curved line represents an object with a changing (non-uniform) acceleration.

The sign of acceleration depends on whether the object is speeding up or slowing down: if an object is speeding up, its acceleration is positive; if it is slowing down, its acceleration is negative — this is also called deceleration. However, acceleration can also be negative simply because the object is accelerating in the negative direction (e.g. speeding up while travelling in the direction taken as negative), so the sign of acceleration alone does not always tell you whether an object is speeding up or slowing down — it must be compared with the sign of the velocity. A rocket lifting off is an example of positive acceleration (speeding up), while a car braking to a stop is an example of negative acceleration (decelerating). The units of acceleration, $\mathrm{m\,s^{-2}}$, follow directly from the definition: it measures how much the velocity (in $\mathrm{m\,s^{-1}}$) changes every second, i.e. $(\mathrm{m\,s^{-1}})\,\mathrm{s^{-1}}$.

The kinematic equations of motion, often called the SUVAT equations, are a set of four equations that describe the motion of any object moving with constant (uniform) acceleration in a straight line. They relate five variables:

• $s$ = displacement
• $u$ = initial velocity
• $v$ = final velocity
• $a$ = acceleration
• $t$ = time interval

The four kinematic equations are:

$v = u + at$

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

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

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

Each equation uses only four of the five variables, so choosing the correct one depends on which quantities are given and which is unknown; the last equation above can in fact be derived by combining two of the others. These equations are provided in the data booklet during exams, so they do not need to be memorised, but recognising when and how to apply them is essential.

Certain key phrases in a question tell you which variables to substitute:

• "Starts from rest" means $u = 0$ (and often $t=0$ marks the start of the motion); this can also be assumed whenever an initial velocity is not stated.
• "Falling due to gravity" means $a = g$, taken as $9.8\,\mathrm{m\,s^{-2}}$ or $9.81\,\mathrm{m\,s^{-2}}$ depending on the precision required. It does not matter which direction is chosen as positive, as long as the choice is applied consistently to every vector quantity in the problem — e.g. if downwards is positive, $a = +9.81\,\mathrm{m\,s^{-2}}$; if upwards is positive, $a = -9.81\,\mathrm{m\,s^{-2}}$.
• "Constant acceleration in a straight line" is the signal that the kinematic equations are the correct tool to use, for example an object falling under uniform gravity with no air resistance.

To apply the kinematic formulae: Step 1 — list every variable given in the question (known and unknown), using the wording of the question to infer any values not explicitly stated (e.g. "starts from rest" gives $u=0$; vertical motion under gravity gives $a = \pm 9.81\,\mathrm{m\,s^{-2}}$). Step 2 — choose the one equation that contains exactly the variables you have listed (e.g. if $s$, $u$, $a$, and $t$ are all known or wanted, use $s = ut + \tfrac12 at^2$). Step 3 — convert all quantities to SI units, substitute them into the chosen equation, and rearrange algebraically to find the answer.

The motion of an object can be analysed using three related types of graph: displacement–time, velocity–time, and acceleration–time graphs. These three graphs are all connected to each other through gradients (which give the next quantity down) and areas under the curve (which give the next quantity up).

Displacement–time graphs. The slope (gradient) at any point equals the velocity at that instant. The $y$-intercept equals the initial displacement. A straight (diagonal) line represents constant velocity; a curved line represents an accelerating object. A positive slope means motion in the positive direction, a negative slope means motion in the negative direction, and a zero slope (horizontal line) means the object is at rest. The area under a displacement–time graph has no physical meaning.

Velocity–time graphs. The slope at any point equals the acceleration at that instant. The $y$-intercept equals the initial velocity. A straight (diagonal) line represents uniform (constant) acceleration; a curved line represents non-uniform acceleration. A positive slope means acceleration in the positive direction, a negative slope means acceleration in the negative direction, and a zero slope (horizontal line) means the object moves at constant velocity. The area under the curve equals the change in displacement.

Acceleration–time graphs. The slope of this graph has no physical meaning. The $y$-intercept equals the initial acceleration. A zero slope (horizontal line) represents an object undergoing constant acceleration. The area under the curve equals the change in velocity.

Acceleration itself can be uniform (a constant value, such as free-fall acceleration due to gravity near Earth's surface) or non-uniform (a changing value, such as an object whose acceleration is itself increasing).

Worked case: a bouncing ball. Ignoring air resistance, a ball bouncing repeatedly reaches the same height each time before striking the ground again, since gravity is the only force acting and is always directed downwards (taking upward as the positive direction, as is conventional). Because the ball reverses direction at its highest and lowest points, its velocity — a vector — changes sign there: it is positive while travelling upward and negative while travelling downward. At point A, the highest point of a bounce: the ball is at its maximum displacement, its velocity is momentarily zero, its velocity changes sign from positive to negative as it changes direction, and its acceleration remains constant at $g$, directed downward throughout. At point B, the lowest point (on the ground): the ball is at its minimum displacement, its velocity changes almost instantaneously from negative to positive while its speed (magnitude of velocity) stays the same, and this abrupt reversal in direction produces a large, momentary acceleration (since acceleration is the change in velocity divided by the — here very short — time over which that change happens).

A projectile is a particle that moves freely (non-powered) under gravity in a two-dimensional plane. Examples include throwing a ball, jumping off a diving board, or hitting a baseball. In standard projectile-motion problems it is assumed that resistance from the air or liquid the object travels through (fluid resistance) is negligible, and that the acceleration due to free fall, $g$, is constant, since the object stays close to the Earth's surface.

A projectile is launched with a resultant velocity $u$ at an angle $\theta$ to the horizontal (for example, a ball thrown from a height, or a cannonball fired from a cannon). Several key quantities describe its motion:

• Time of flight — the total time the projectile spends in the air. For a projectile that lands back at its launch height, the time to reach maximum height is exactly half of the total time of flight.
• Maximum height — the height at which the projectile is momentarily at rest vertically, i.e. where its vertical velocity component is zero. This occurs when half of the total time of flight has elapsed (for a projectile launched and landing at the same height).
• Range — the total horizontal distance travelled by the projectile.

Horizontal and vertical components. The trajectory of a projectile consists of a horizontal component and a vertical component, and these two components are entirely independent of each other. Displacement, velocity, and acceleration each need to be evaluated separately in the horizontal and vertical directions using the kinematic (SUVAT) equations:

The resultant launch velocity $u$ at angle $\theta$ to the horizontal is split into its two components using trigonometry, treating $u$ as the hypotenuse of a right-angled triangle:

$u_{\text{vertical}} = u\sin\theta \qquad u_{\text{horizontal}} = u\cos\theta$

To solve a projectile problem, resolve the launch velocity into these two components, then set up separate SUVAT tables for the horizontal motion (constant velocity, $a=0$) and the vertical motion (constant acceleration $a = \pm g$), solving each using the standard kinematic equations. Common scenarios you may need to solve include: vertical projection above the horizontal, vertical projection below the horizontal, purely horizontal projection, and projection at an angle (the most common case).

Fluid resistance refers to the effect that gases and liquids have on the motion of a body moving through them. When an object moves through a fluid, resistive forces act on it; these are known as viscous drag. Viscous drag — also called air resistance when the fluid is air — is a type of friction, and like all frictional forces it: always acts in the direction opposite to the object's motion; never speeds an object up or starts it moving; always slows an object down or keeps it moving at constant speed; and always transfers energy away from the object to the surroundings. A related force, lift, acts perpendicular to the fluid flow (for example, as an aeroplane's wings push down on the air, the air pushes back up on the wings by Newton's third law, generating lift); drag opposes the direction of motion (thrust), while lift opposes weight. A key feature of drag forces is that they increase as the speed of the object increases.

Effect on projectile motion. When fluid resistance is not negligible, it affects a projectile's time of flight, horizontal velocity, horizontal acceleration, range, and the shape of its trajectory. Air resistance is generally the dominant frictional force acting on a projectile, and it decreases the horizontal component of the projectile's velocity throughout the flight. As a result, compared with an identical situation with no air resistance (as in a vacuum):

Both the range and the maximum height of the projectile decrease relative to the no-resistance case, and its flight time shortens accordingly, since it covers a smaller range and reaches a lower maximum height. The path also stops being a symmetric parabola: with air resistance present, the descending part of the trajectory is steeper than the ascending part. Because of these effects, the launch angle and speed of a projectile are chosen differently depending on the goal and the presence of air resistance: in sports such as the long jump or javelin, an optimum angle (accounting for air resistance) is chosen to maximise range, whereas in activities such as gymnastics or ski jumping, the initial vertical velocity is maximised to achieve the greatest height and longest flight path.

For a body in free fall in a vacuum, the only force acting on it is its weight, so its acceleration is $g$, due to gravity alone, and stays constant throughout the fall. When the same body falls through a real fluid (a gas or liquid), a frictional force from fluid resistance (viscous drag) also acts on it, and this drag force increases as the body's speed increases.

By Newton's second law ($F = ma$), as the drag force grows, the resultant (net) force on the body — weight minus drag — gets smaller, so the body's acceleration decreases. Once the viscous drag force becomes exactly equal in magnitude to the body's weight, the resultant force on the body is zero, so it stops accelerating and continues falling at a constant velocity. This constant falling velocity is called the terminal velocity (or terminal speed). Terminal velocity can be reached by an object falling through either a gas or a liquid.

On a velocity–time graph of a body falling to terminal velocity, since acceleration equals the gradient of the graph: the gradient (acceleration) is steepest at the start of the fall and continuously decreases as speed increases, and it becomes zero exactly when the terminal velocity is reached — shown on the graph as the curve flattening out to a horizontal line. For a skydiver who later opens a parachute, the sudden large increase in drag once the parachute deploys causes a rapid deceleration to a new, lower terminal velocity, which reduces the impact of landing; on the velocity-time graph this appears as a steep drop in velocity followed by the curve flattening again at this lower constant speed.