UK A-Level: Kinematics

Kinematics is the language of motion, allowing you to describe how objects move without needing to explain why. Mastering it is essential for progressing into dynamics and the broader study of physics, from designing safer cars to planning space missions. In your A-Level studies, you’ll move beyond simple formulas to analyse motion with mathematical precision, tackling both constant and variable acceleration.

Graphs as the Language of Motion

The journey into kinematics begins with interpreting graphs. A displacement-time ($s$-$t$) graph tells a story about position. The gradient (slope) of this graph at any point is the object's velocity. A straight line indicates constant velocity, a curve indicates changing velocity. Crucially, the sign of the gradient matters: a positive gradient means motion in the positive direction, a negative gradient means motion back toward the origin.

The velocity-time ($v$-$t$) graph provides deeper insight. Here, the gradient represents acceleration. A horizontal line means constant velocity (zero acceleration), while a sloping line means constant, non-zero acceleration. More powerfully, the area under a velocity-time graph represents the displacement. For motion below the time axis (negative velocity), the area counts as negative displacement. This relationship is foundational: differentiation connects displacement to velocity ($v=ds/dt$), and integration connects velocity back to displacement ($s=\int vdt$).

The SUVAT Equations: Toolkit for Constant Acceleration

When acceleration is constant, a powerful set of five equations, known as the SUVAT equations, apply. They are derived from the definitions of acceleration and the area under a $v$-$t$ graph. The variables are: $s$ (displacement), $u$ (initial velocity), $v$ (final velocity), $a$ (acceleration), and $t$ (time).

The core equations are: $$v=u+at$$ $$s=\frac{(u+v)t}{2}$$ $$s=ut+\frac{1}{2}a{t}^2$$ $$v^2=u^2+2as$$

To use them effectively, you must first establish a positive direction and ensure all vector quantities ($s$, $u$, $v$, $a$) have the correct sign relative to it. Then, list the known variables and the one you need to find, choosing the equation that connects them without the unknown variable you don't have.

*Example: A car accelerates from rest at $3{\text{ m s}}^{-2}$ for 5 seconds. Find its displacement.*

1. Positive direction = direction of travel.
2. Known: $u=0$, $a=3$, $t=5$. Unknown: $s$.
3. Use $s=ut+\frac{1}{2}a{t}^2$.
4. $s=(0\times 5)+\frac{1}{2}(3)(5^2)=37.5\text{ m}$.

Calculus and Variable Acceleration

Real-world motion often involves changing acceleration, such as a car with a powerful engine. Here, the SUVAT equations fail, and you must use calculus. The relationships are defined differentially:

Velocity is the rate of change of displacement: $v=\frac{ds}{dt}$. Acceleration is the rate of change of velocity: $a=\frac{dv}{dt}$.

Conversely, you integrate to move in the opposite direction: $$v=\int adt$$ $$s=\int vdt$$

You will often be given an expression for $a$ in terms of $t$ (e.g., $a=2t+1$). To find velocity, integrate with respect to time, adding a constant of integration which you determine using an initial condition (e.g., $u=5$ when $t=0$). To then find displacement, integrate the velocity function.

*Example: A particle moves such that $a=4t$, with $u=2$ and $s=0$ at $t=0$. Find $s$ when $t=3$.*

1. Find $v$: $v=\int 4tdt=2t^2+C$. Since $v=2$ when $t=0$, $C=2$. So $v=2t^2+2$.
2. Find $s$: $s=\int (2t^2+2)dt=\frac{2}{3}{t}^3+2t+K$. Since $s=0$ when $t=0$, $K=0$.
3. Evaluate at $t=3$: $s=\frac{2}{3}(27)+2(3)=18+6=24\text{ m}$.

Vertical Motion Under Gravity

A prime application of constant-acceleration kinematics is motion under gravity near the Earth's surface. Ignoring air resistance, all objects experience the same constant acceleration due to gravity, $g$. Its magnitude is approximately $9.8{\text{ m s}}^{-2}$. The critical step is defining your sign convention at the start and sticking to it rigorously.

Typically, upwards is taken as positive. Consequently:

• Acceleration is always $a=-g=-9.8{\text{ m s}}^{-2}$.
• An initial upward throw has positive $u$.
• An object dropped from rest has $u=0$.
• At the highest point of a trajectory, velocity $v=0$, but acceleration is still $-9.8{\text{ m s}}^{-2}$.

*Example: A ball is thrown upwards at $15{\text{ m s}}^{-1}$ from ground level. Find its maximum height.*

1. Positive direction = upwards. Known: $u=15$, $v=0$ (at max height), $a=-9.8$. Unknown: $s$.
2. Use $v^2=u^2+2as$.
3. $0^2=15^2+2(-9.8)s$
4. $0=225-19.6s⟹s=225/19.6\approx 11.5\text{ m}$.

Common Pitfalls

1. Sign Convention Errors: The most frequent mistake is mixing signs within a problem. Decide on positive direction (e.g., right or up) before writing any equations. If motion occurs in the opposite direction, its displacement, velocity, and acceleration must be negative. In vertical motion, never treat $g$ as $+9.8$ if up is positive.

1. Misinterpreting Graph Areas: Remember, area under a velocity-time graph gives displacement. Area under an acceleration-time graph gives the change in velocity. Never find area on a displacement-time graph for kinematic quantities.

1. Using SUVAT with Variable Acceleration: A sure path to an incorrect answer is applying $s=ut+\frac{1}{2}a{t}^2$ when acceleration is not constant. If acceleration is expressed as a function of time ($a=4t$, $a=\sin t$), you must use calculus methods.

1. Confusing Velocity and Speed at Turning Points: In vertical motion, when an object reaches its peak height, its velocity is zero. Its acceleration, however, is still $-9.8{\text{ m s}}^{-2}$ acting downwards. It is not momentarily stationary with no forces acting.

Summary

• The gradient of a displacement-time graph gives velocity; the gradient of a velocity-time graph gives acceleration. The area under a velocity-time graph gives displacement.
• The five SUVAT equations apply only for motion with constant acceleration. Success depends on a strict, consistent sign convention for all vector quantities.
• For variable acceleration, you must use calculus: differentiate displacement to get velocity, and velocity to get acceleration; integrate acceleration to get velocity, and velocity to get displacement.
• In vertical motion under gravity, acceleration is constant at $a=-g$ (assuming upwards is positive). At the peak of a trajectory, velocity is zero but acceleration remains $-g$.
• Always translate a problem into a clear diagram with your defined positive direction before writing any equations. This simple step prevents most sign-related errors.