Kinematics

All motion is measured relative to a chosen origin and positive direction, so the first step in any kinematics problem is fixing that frame.

Distance is a scalar: the total length of the path actually travelled, which can only stay the same or increase. Displacement is a vector: the straight-line change in position from start to finish, with both a magnitude and a direction, and it can shrink back to zero if the object returns to where it began.

Speed and velocity follow the same split. Average speed is total distance divided by total time, while average velocity is total displacement divided by total time. Instantaneous velocity is the rate of change of displacement at one moment: the gradient of the displacement-time graph at that instant, given over a vanishingly short time interval by:

$$v = \frac{\Delta s}{\Delta t}$$

Its magnitude is the instantaneous speed.

A concrete example makes the distinction sharp: an athlete who runs one complete $400\ \text{m}$ track lap in $50\ \text{s}$ covers a distance of $400\ \text{m}$ (average speed $8.0\ \text{m s}^{-1}$) but has zero displacement, so the average velocity is zero.

You must be able to:

• define each quantity precisely
• classify it as scalar or vector
• attach correct signs to one-dimensional motion
• calculate average values from journey data

Acceleration is the rate of change of velocity at one moment: the gradient of the velocity-time graph at that instant, given over a vanishingly short time interval by:

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

Because velocity is a vector, an object accelerates whenever its speed or its direction changes. A negative acceleration does not automatically mean slowing down; it means the acceleration vector points in the negative direction, which speeds the object up if it is already moving that way.

Motion graphs encode this information geometrically:

• On a displacement-time graph, the gradient gives velocity; a curved line means the velocity is changing.
• On a velocity-time graph, the gradient gives acceleration and the area between the line and the time axis gives displacement, with area below the axis counting as negative.
• On an acceleration-time graph, the area gives the change in velocity.

For example, a car that accelerates uniformly from rest, cruises, then brakes produces a velocity-time graph with a rising straight line, a horizontal section, and a falling line; the total area under all three sections is the distance travelled.

You should be able to:

• sketch each graph type for a described motion
• convert between graph types
• read gradients with a tangent for curved lines
• estimate areas under non-linear curves by counting grid squares

When acceleration is constant along a straight line, five quantities fully describe the motion:

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

The four SUVAT equations each link four of these five, so the practical skill is to list the three quantities you know, identify the one you want, and pick the equation that omits the quantity you neither know nor need.

For example, the equation that avoids time is:

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

while the equation that avoids final velocity is:

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

The non-negotiable condition is constant acceleration: the equations fail for a car whose braking force varies or for a falling object with air resistance. Free fall near the Earth's surface (with resistance ignored) is the classic valid case, with $a = 9.8\ \text{m s}^{-2}$ directed downwards.

A clear sign convention is essential. Choosing upwards as positive for a thrown ball makes $u$ positive, $a = -9.8\ \text{m s}^{-2}$, and a displacement below the launch point negative.

You must be able to select and rearrange the appropriate equation, apply consistent signs, and solve multi-stage problems by treating each constant-acceleration stage separately, carrying the final velocity of one stage forward as the initial velocity of the next.

A projectile is an object moving freely under gravity alone after launch. The central idea is independence of components: the horizontal motion has zero acceleration ($a_x = 0$), so the horizontal velocity stays constant, while the vertical motion is uniform acceleration with $a_y = -g$.

The two are connected only through the shared time of flight $t$, which is why the trajectory is a parabola.

The standard method is to resolve the launch velocity into components:

$$u_x = u\cos\theta, \qquad u_y = u\sin\theta$$

Then treat each direction with its own SUVAT analysis. At the highest point the vertical velocity is zero but the horizontal velocity is unchanged, so the projectile is still moving. For a ball rolled horizontally off a table, the fall time depends only on the height, and the landing distance is just the horizontal speed multiplied by that time.

With air resistance included, the path becomes asymmetric: the maximum height, range, and final speed are all reduced, and the descent is steeper than the ascent.

You should be able to calculate time of flight, maximum height, and range for level or elevated launches, and sketch how resistance distorts the ideal parabola.

When an object moves through a fluid, it collides with fluid molecules, generating a drag force $F_d$ that opposes its motion. This drag force typically depends on the velocity of the object: $F_d \propto v$ at low speeds (laminar flow) or $F_d \propto v^2$ at high speeds (turbulent flow).

As an object falls from rest under gravity, the net downward force is:

$$F_{\text{net}} = mg - F_d$$

Because velocity increases, $F_d$ grows, which continuously decreases the net acceleration $a = \frac{mg - F_d}{m}$. This decrease in acceleration is non-linear and continues until $F_d = mg$, where the net force becomes zero and the object moves at a constant terminal speed.

A skydiver illustrates the whole story:

• immediately after leaving the aircraft the acceleration is $9.8\ \text{m s}^{-2}$
• the velocity-time graph then curves over as drag builds
• the diver levels off at a terminal speed
• opening the parachute increases the drag enormously, decelerating the diver to a new, much lower terminal speed for landing

Because the acceleration changes throughout, SUVAT equations cannot be applied to this motion. You should be able to explain each stage using force arguments, sketch the corresponding velocity-time and acceleration-time graphs (initial gradient $g$, approaching a horizontal asymptote), and state the terminal-speed condition.