IB Physics: Mechanics - Kinematics

Kinematics is the language of motion, providing the fundamental tools to describe how objects move through space and time. In IB Physics, a strong command of kinematics is non-negotiable—it forms the essential groundwork for understanding dynamics, energy, and the more complex topics that follow.

Defining the Core Quantities: Displacement, Velocity, and Acceleration

Before solving complex problems, you must internalize the precise, physics-based definitions of three core quantities. Displacement is a vector quantity describing an object's change in position from its starting point. It is "as the crow flies" distance with a specific direction, measured in metres (m). This is distinct from distance travelled, which is a scalar (directionless) total of the path length.

Velocity is the rate of change of displacement. Its formal definition is $v=\frac{\Delta s}{\Delta t}$, where $\Delta s$ is the change in displacement. Because displacement is a vector, velocity is also a vector, possessing both magnitude (speed) and direction, measured in metres per second (ms${}^{-1}$). Acceleration is the rate of change of velocity, defined as $a=\frac{\Delta v}{\Delta t}$. Crucially, an object is accelerating if its velocity changes in any way—whether it's speeding up, slowing down, or changing direction. Acceleration is measured in metres per second squared (ms${}^{-2}$).

A common and vital conceptual step is interpreting the sign of these vectors in one-dimensional motion. In a chosen positive direction (e.g., right or upwards), a positive velocity means motion in that direction. A positive acceleration means the velocity is becoming more positive. This means if velocity is positive, a positive acceleration increases speed. However, if velocity is negative (motion opposite the positive direction), a positive acceleration actually decreases the speed, as it makes the negative number less negative, bringing it closer to zero.

Graphical Analysis: The Story in the Slopes and Areas

Kinematic graphs are not just plots; they are rich, interconnected narratives of motion. Each graph type provides specific information, and the connections between them are governed by calculus principles of gradients and areas.

A displacement-time (s-t) graph tells you where the object is. The gradient (slope) of the tangent to the curve at any point gives the instantaneous velocity. A curved line indicates changing velocity, and thus acceleration.

A velocity-time (v-t) graph is arguably the most powerful. Its gradient gives the instantaneous acceleration. The area under the v-t graph, between two times, gives the change in displacement over that interval. This area is algebraic—areas below the time axis (negative velocity) count as negative displacement.

An acceleration-time (a-t) graph's gradient is rarely used in IB. However, the area under an a-t graph gives the change in velocity. A horizontal line on an a-t graph indicates constant, or uniform, acceleration.

Example: Consider a v-t graph showing a straight line sloping downwards from a positive velocity to zero. The negative gradient tells you acceleration is constant and negative (deceleration). The area under this triangle gives the total displacement travelled before stopping.

The SUVAT Equations: Solving for Uniform Acceleration

When acceleration is uniform (constant in both magnitude and direction), five key variables link an object's motion: displacement ($s$), initial velocity ($u$), final velocity ($v$), acceleration ($a$), and time elapsed ($t$). The SUVAT equations are a set of four relationships between these five variables. You must know them by heart:

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

To apply these effectively, follow a disciplined process: 1) Define a positive direction. 2) Write down the five SUVAT symbols ($s,u,v,a,t$). 3) Fill in the values you know from the question, including signs based on your defined direction. 4) Identify the unknown you need to find. 5) Choose the equation that includes the known variables and the desired unknown. This methodical approach prevents errors.

Worked Example: A ball is thrown vertically upwards from ground level with an initial speed of $20$ ms${}^{-1}$. Taking $g=10$ ms${}^{-2}$ and upwards as positive, find its maximum height. Step 1: Positive direction is upwards. Step 2 & 3: List knowns: $s=?$ (what we want), $u=+20$ ms${}^{-1}$, $v=0$ ms${}^{-1}$ (at the top of flight), $a=-10$ ms${}^{-2}$, $t=?$ (not needed). Step 4 & 5: We know $u,v,a$ and want $s$. Use $v^2=u^2+2as$. $$0^2=20^2+2(-10)s$$ $$0=400-20s$$ $$20s=400$$ $$s=20\text{m}$$

Projectile Motion: Two-Dimensional Analysis

Projectile motion is the classic application of two-dimensional kinematics. The key insight is that horizontal and vertical motions are completely independent. The only force acting (in ideal cases) is gravity, which causes a constant downward acceleration $a_y=-g$. There is no horizontal acceleration ($a_x=0$).

To solve any projectile problem, you must resolve the initial velocity vector into its horizontal ($u_x$) and vertical ($u_y$) components. If a projectile is launched with speed $u$ at an angle $\theta$ above the horizontal, then: $$u_x=u\cos \theta$$ $$u_y=u\sin \theta$$

You then treat the problem as two separate, simultaneous one-dimensional motions. For the horizontal motion, velocity is constant ($v_x=u_x$), so you use $s_x=u_{x}t$. For the vertical motion, acceleration is uniform ($a_y=-g$), so you use the SUVAT equations with $a=-g$.

Key Results: For a projectile launched and landing on the same horizontal level, its time of flight is $t=\frac{2u\sin \theta }{g}$, its maximum height is $H=\frac{u^2{\sin }^2\theta }{2g}$, and its horizontal range is $R=\frac{u^2\sin 2\theta }{g}$.

Relative Motion

Relative motion analysis asks: "How does object A appear to be moving from the perspective of object B?" The fundamental equation for velocity is: $${\vec{v}}_{A\text{ relative to }B}={\vec{v}}_A-{\vec{v}}_B$$ where all velocities are measured relative to the same frame (usually the ground). For example, if two cars are moving in the same direction on a highway, a car traveling at $30$ ms${}^{-1}$ relative to the ground will appear to be moving backwards at $5$ ms${}^{-1}$ from the perspective of a car traveling at $35$ ms${}^{-1}$ in the same direction, because $30-35=-5$ ms${}^{-1}$.

This concept is crucial for problems involving crossing rivers (where the boat's velocity relative to the water must be combined with the water's velocity relative to the earth) or for aircraft navigating in wind.

Common Pitfalls

1. Confusing velocity and speed, or displacement and distance. Remember: velocity and displacement are vectors and depend on direction. A car driving in a circle at constant speed has a changing velocity and zero displacement after one lap, but a positive distance travelled.

Correction: Always ask: "Does the question care about direction?" If it asks for "how far from the start," it likely wants displacement. If it asks for "total ground covered," it wants distance.

1. Misinterpreting graph gradients and areas. A common error is to think the gradient of an s-t graph gives speed (it gives velocity, which includes sign), or that the area under an a-t graph gives displacement (it gives change in velocity).

Correction: Practice sketching one type of graph from another. For example, given a v-t graph with a straight line, sketch the corresponding s-t graph (it will be a parabola).

1. Mishandling signs in SUVAT equations. Using incorrect signs for $g$, or mixing positive and negative displacements/velocities inconsistently, is the primary source of algebraic error.

Correction: Before writing any equation, explicitly state your positive direction (e.g., "Upwards is positive"). Then assign a + or - sign to every vector quantity ($s,u,v,a$) based on that convention.

1. Treating projectile motion as one combined motion. Students often try to use a single SUVAT equation for the diagonal path, which fails because acceleration is not along the launch direction.

Correction: Resolve, resolve, resolve. Immediately break the initial velocity into $u_x$ and $u_y$. Write two separate lists of knowns for the horizontal and vertical motions. The shared variable that links them is time ($t$).

Summary

• Kinematics describes motion through the vector quantities displacement, velocity (rate of change of displacement), and acceleration (rate of change of velocity).
• Graphs are interconnected: on a velocity-time graph, the gradient equals acceleration and the area under the curve equals change in displacement.
• The SUVAT equations ($v=u+at$, $s=ut+\frac{1}{2}a{t}^2$, $v^2=u^2+2as$, $s=\frac{(u+v)t}{2}$) are the toolkit for solving problems involving uniform acceleration. A disciplined sign convention is critical.
• Projectile motion is analyzed by resolving the initial velocity into independent horizontal (constant velocity) and vertical (constant acceleration) components.
• Relative velocity is found by vector subtraction: ${\vec{v}}_{A/B}={\vec{v}}_A-{\vec{v}}_B$, allowing you to analyze motion from different frames of reference.