Do these equations work for objects falling under gravity?

Yes! Free fall is just 1D kinematics with a constant downward acceleration. You just use $a = -9.8 m/s^2$ (assuming up is positive) and swap $\Delta x$ for $\Delta y$ since the object is moving vertically.

What if I only have two known variables?

If you only see two numbers in the problem, look closely for hidden words. Phrases like 'drops', 'comes to a stop', or 'starts from rest' are secret ways of giving you a velocity of zero.

Can I use these equations if the acceleration is changing?

No. These specific four kinematics equations only work when acceleration is constant. If acceleration is changing over time, you have to use calculus.

How do you pick the right kinematics equation?

To pick the right kinematics equation, you simply list out the variables you know from the word problem and the one specific variable you are trying to find. Every standard kinematics problem involves five core variables, but each of the four main equations only uses four of them. This means every equation has exactly one variable completely missing from it.

Think of it like playing a game of Guess Who. If you know the mystery person isn't wearing a hat, you flip down all the characters wearing hats to narrow down your choice. In physics, if a problem doesn't mention time and doesn't ask you to find time, you simply pick the one equation that doesn't have time in it.

The Big Five Variables

Before picking an equation, you need to know the cast of characters. In 1D kinematics, there are five variables we care about. These are displacement ( $\Delta x$ ), initial velocity ( $v_0$ ), final velocity ( $v$ ), acceleration ( $a$ ), and time ( $t$ ). Every kinematics problem will give you three of these, ask you for a fourth, and completely ignore the fifth.

The Four Equations and Their Missing Pieces

Because each equation only connects four variables, the easiest way to memorize them is by remembering what they leave out. The first equation is $v = v_0 + at$ (missing displacement). The second is $\Delta x = v_0 t + \frac{1}{2}at^2$ (missing final velocity). The third is $v^2 = v_0^2 + 2a\Delta x$ (missing time). The fourth is $\Delta x = \frac{v + v_0}{2}t$ (missing acceleration). If you figure out which variable is the "who cares" variable, your choice is made.

A Simple Three-Step Process

First, read the problem and write down your "Knowns" (the three numbers given). Look out for hidden knowns, like "starts from rest" which means $v_0 = 0$ . Second, write down your "Unknown" (the one thing you need to find). Third, look at the one variable left over that you didn't write down. That is your "who cares" variable. Pick the equation that is missing that exact variable.

Where Students Slip Up

The most common mistake isn't picking the wrong equation, but getting the positive and negative signs wrong once the equation is picked. Kinematics equations rely on vectors, which have direction. If an object is moving right but slowing down, its velocity is positive but its acceleration must be negative. Always define which way is positive before you start plugging numbers into your chosen equation.

Worked through

A car accelerates from rest at a constant rate of $3.0 m/s^2$ over a distance of $150 m$ . What is the final velocity of the car?

Step 1: Write down the knowns. The car starts from rest, so $v_0 = 0 m/s$ . The acceleration is $a = 3.0 m/s^2$ . The displacement is $\Delta x = 150 m$ .

Step 2: Write down the unknown. We want to find the final velocity, so $v = ?$

Step 3: Identify the missing variable. We have $v_0$ , $a$ , $\Delta x$ , and $v$ . The variable we don't know and don't care about is time ( $t$ ).

Step 4: Pick the equation missing time. That is $v^2 = v_0^2 + 2a\Delta x$ .

Step 5: Plug in the numbers and solve. $v^2 = (0)^2 + 2(3.0)(150)$ $v^2 = 900$ $v = 30 m/s$ The car's final velocity is $30 m/s$ .

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Where this comes from: OpenStax Physics: Chapter 3 (Kinematics) · Khan Academy: One-dimensional motion