Kinematics: SUVAT Equations and Projectile Motion

Mastering kinematics is essential for understanding how objects move in our world, from a car accelerating on a highway to a basketball arcing toward the hoop. For IB Physics, this topic forms the bedrock for more advanced mechanics, requiring you to derive and apply core equations to solve complex, real-world motion problems. By the end of this guide, you will be able to analyze any scenario involving uniform acceleration and two-dimensional trajectories with confidence.

Foundations: Deriving the SUVAT Equations

Kinematics describes motion without considering the forces that cause it. The cornerstone of this description for uniform acceleration—constant acceleration—is the set of SUVAT equations. The acronym SUVAT comes from the five key variables: S (displacement), U (initial velocity), V (final velocity), A (acceleration), and T (time). You must be comfortable deriving these equations from fundamental definitions.

The starting point is the definition of acceleration: $a=\frac{v-u}{t}$. A simple rearrangement gives our first SUVAT equation: $$v=u+at$$

The second equation comes from considering average velocity. Under uniform acceleration, the average velocity is the mean of the initial and final velocities: $v_{avg}=\frac{u+v}{2}$. Since displacement is average velocity multiplied by time, we substitute our expression for $v$ from the first equation: $$s=v_{avg}\times t=\left(\frac{u+v}{2}\right)t$$ Substituting $v=u+at$ gives: $$s=\left(\frac{u+(u+at)}{2}\right)t=\left(\frac{2u+at}{2}\right)t$$ This simplifies to our second key equation: $$s=ut+\frac{1}{2}a{t}^2$$

The third key equation eliminates time ($t$). Start with $v=u+at$ and square both sides: $v^2=(u+at{)}^2=u^2+2uat+a^2{t}^2$. Notice that $2uat+a^2{t}^2=2a(ut+\frac{1}{2}a{t}^2)$. The term in parentheses is simply displacement $s$. Therefore, we arrive at: $$v^2=u^2+2as$$

These three equations, along with the definitions of average velocity and acceleration, are your toolkit for any one-dimensional uniformly accelerated motion problem. Your first step should always be to list your known SUVAT variables and identify the unknown you need to find.

Analyzing Motion with Graphs

Graphical analysis provides a powerful visual intuition for kinematics. For displacement-time ($s$-$t$) graphs, the gradient (slope) at any point represents the instantaneous velocity. A curved line indicates changing velocity, and thus acceleration. For velocity-time ($v$-$t$) graphs, the gradient represents acceleration, while the area under the graph represents displacement.

Consider a car that accelerates uniformly from rest at $4{\text{ m s}}^{-2}$ for 5 seconds, then travels at constant velocity for 10 seconds. Its $v$-$t$ graph would be a straight line sloping upward from 0 to 20 m/s over the first 5 seconds (area = a triangle: displacement = $\frac{1}{2}\times 5\times 20=50\text{ m}$), followed by a horizontal line at 20 m/s for 10 seconds (area = a rectangle: displacement = $20\times 10=200\text{ m}$). The total displacement is the sum of these areas: 250 m. This method is often quicker than using equations when motion changes in distinct segments. Conversely, a horizontal line on an $s$-$t$ graph (zero gradient) indicates an object at rest, while a straight line with a positive gradient indicates constant velocity.

Projectile Motion: Two-Dimensional Analysis

Projectile motion is the classic application of two-dimensional kinematics. The central principle is the independence of horizontal and vertical motion components. Gravity only affects the vertical component; there is no horizontal acceleration (assuming air resistance is negligible). Therefore, you must resolve the initial velocity ($u$) into horizontal ($u_x$) and vertical ($u_y$) components using trigonometry: $$u_x=u\cos \theta \text{and}{u}_y=u\sin \theta$$ where $\theta$ is the launch angle from the horizontal.

The horizontal motion is simple: constant velocity. $$s_x=u_{x}t$$

The vertical motion is uniformly accelerated motion, with acceleration $a_y=-g$ (where $g=9.8{\text{ m s}}^{-2}$, negative as we typically define upward as positive). We apply the SUVAT equations vertically: $$v_y=u_y-gt$$ $$s_y=u_{y}t-\frac{1}{2}g{t}^2$$

These two sets of equations are linked by time ($t$), which is the same for both dimensions. To solve a typical problem, like finding the range (total horizontal distance) of a projectile launched from level ground:

1. Resolve the initial velocity.
2. Analyze the vertical motion to find the time of flight. When the projectile lands, $s_y=0$. The equation $0=u_{y}t-\frac{1}{2}g{t}^2$ gives $t=0$ (launch) and $t=\frac{2u_y}{g}$ (landing).
3. Use this total time in the horizontal motion equation: $s_x=u_x\times \frac{2u_y}{g}$.

This results in a symmetric parabolic trajectory. The maximum height is achieved when the vertical velocity becomes zero ($v_y=0$), which you can find using $v_y^2=u_y^2-2g{s}_y$.

Common Pitfalls

1. Mixing Horizontal and Vertical Components: The most frequent error is using a horizontal value in a vertical SUVAT equation, or vice versa. Velocity, displacement, and acceleration are vector components. You must keep the $x$ and $y$ directions separate until the final step where you might combine them vectorially. For example, initial velocity $u$ is not a valid entry for a vertical SUVAT equation; you must use $u_y$.

1. Sign Convention Errors with Gravity: Deciding on a positive direction at the start of a problem and sticking to it is crucial. If you define upward as positive, then acceleration due to gravity is $a_y=-9.8{\text{ m s}}^{-2}$. A projectile launched upward will have a positive $u_y$, but its velocity will become less positive, pass through zero, and then become negative. Displacement can be positive (above launch point) or negative (below). Inconsistent signs are a primary source of calculation mistakes.

1. Misinterpreting "Initial" and "Final": In multi-stage problems, the "final" velocity of one stage becomes the "initial" velocity for the next. For instance, for a ball thrown upwards, at the maximum height its vertical velocity is zero. This is the "final" velocity for the upward journey and the "initial" velocity for the downward journey. However, its horizontal velocity remains constant throughout if air resistance is ignored.

1. Forgetting the Independence of Motion: It's easy to intuitively think that an object with a greater horizontal velocity will stay in the air longer. This is false. The time of flight in projectile motion from level ground is determined solely by the vertical component of motion: $t=\frac{2u_y}{g}$. Two objects launched with the same $u_y$ but different $u_x$ will land at the same time, though at very different horizontal distances.

Summary

• The SUVAT equations ($v=u+at$, $s=ut+\frac{1}{2}a{t}^2$, $v^2=u^2+2as$) are derived from definitions of acceleration and average velocity, and they solve all one-dimensional uniform acceleration problems.
• On a velocity-time graph, the gradient equals acceleration and the area under the graph equals displacement. On a displacement-time graph, the gradient equals velocity.
• Projectile motion is analyzed by resolving motion into independent horizontal (constant velocity) and vertical (uniform acceleration due to gravity) components. The key equations are $s_x=u_{x}t$ and $s_y=u_{y}t-\frac{1}{2}g{t}^2$.
• The time of flight for a projectile launched from and landing at the same vertical height depends only on the initial vertical velocity and gravity.
• Always maintain a consistent sign convention (e.g., up is positive) and never mix horizontal and vertical vector components within the same kinematic equation.
• Mastering these concepts allows you to model and predict the motion of any object under constant acceleration, a fundamental skill for IB Physics.