The Kinematic Equations: Predicting Motion Before It Happens

A car accelerates from rest at a steady rate and covers 100 m in 5 seconds. How fast is it going at the end? A tempting first guess is to divide: $\dfrac{100\text{ m}}{5\text{ s}} = 20$ m/s, and call that the final speed. That guess is wrong — but understanding why it's wrong is exactly what today's equations are built to explain.

A few lessons back, Galileo rolled bronze balls down ramps and discovered that for constant acceleration starting from rest, distance is proportional to time squared: $\Delta x \propto t^2$. In modern notation, that becomes $\Delta x = \tfrac{1}{2}at^2$, extended to handle a nonzero starting velocity as $\Delta x = v_it + \tfrac{1}{2}at^2$. Plugging the car's numbers into this equation and solving gives an acceleration of 8 m/s². From there, a second equation — built directly from the definition of acceleration itself, $a = \dfrac{\Delta v}{\Delta t}$, rearranged into $v = v_i + at$ — gives a final velocity of 40 m/s, not 20. That earlier guess of "20 m/s" wasn't wrong by accident: for constant acceleration starting from rest, the final velocity is always exactly double the average velocity. The 20 m/s from dividing distance by time really is a correct number — it's just the average velocity over the whole 5 seconds, not the final one.

Two more equations round out the toolkit, each useful when a different piece of information is missing: $v^2 = v_i^2 + 2a\Delta x$ (handy when time isn't known) and $\Delta x = \tfrac{1}{2}(v_i + v)t$ (handy when acceleration isn't known). Altogether there are five variables that describe this kind of motion — initial velocity, final velocity, acceleration, displacement, and time — and each of the four equations relates exactly four of them, leaving exactly one out. That gives a simple, reliable strategy for solving any problem: look at which variable is missing from the given information, and reach for the one equation that doesn't need it.

These equations only work under one condition: acceleration has to be constant the entire time. That assumption — called uniformly accelerated motion — holds for a surprising number of real situations, including the two big applications coming next: free fall and projectile motion, both of which are just these same four equations with gravity plugged in as the acceleration.

Try It: Drag Race

Use the kinematic equations to hit exact times, speeds, and stopping distances: Kinematics Castle: Drag Race.

Try It: Catch the Cop

Set two positions equal to catch a speeder: kinematics for two moving objects. Kinematics Castle: Catch the Cop

Try It: Traffic Light Dash

Combine reaction time and braking to find a car's stopping distance. Kinematics Castle: Traffic Light Dash

Videos

Introduction to Uniformly Accelerated Motion with Examples of Objects in UAM — Flipping Physics

Practice Problems

An object starts from rest and accelerates at 5 m/s² for 3 seconds. Find its final velocity.
An object starts at 2 m/s and accelerates at 4 m/s² for 3 seconds. Find the distance it travels.
An object speeds up from 10 m/s to 30 m/s over 4 seconds. Find (a) its acceleration and (b) the distance it traveled during this time.
An object starts from rest and accelerates uniformly, covering 54 m in the process, ending with an acceleration of 3 m/s². Find (a) its final velocity and (b) the time it took.
A ball is thrown with an initial velocity of 15 m/s and decelerates at 3 m/s² (no gravity yet — just a general deceleration). Find (a) how long it takes to come to a stop and (b) how far it travels before stopping.
A train starts from rest and accelerates uniformly at 1.5 m/s² for 20 seconds. a) Find the train's final velocity. b) Find the distance the train traveled. c) If the train instead accelerated for 40 seconds (double the time) at the same rate, justify why the distance traveled would be MORE than double your answer to (b) — not just double it. d) Predict what would happen to the distance traveled (in the original 20 seconds) if the acceleration were doubled instead of the time. Is this relationship different from part (c)? Explain.
A classmate claims: "If two objects both start from rest at the same position and the same instant, and both undergo constant acceleration, the one with the greater acceleration will always have traveled farther after any given amount of time has passed." Evaluate this claim — is it true or false? Justify your answer using the equation $\Delta x = \tfrac{1}{2}at^2$.