Kinematic Equations

The video lesson for this topic is coming soon.

Essential Question

How can we predict motion when acceleration is constant?

This lesson brings together displacement, velocity, and acceleration into four equations that can answer almost any question about steady speeding up or slowing down. By the end of this lesson, you should be able to choose the right kinematic equation and solve for the unknown.

The Five Variables

d = displacement, in meters (m)
v_i = initial velocity, in m/s
v_f = final velocity, in m/s
a = acceleration, in m/s²
t = time, in seconds (s)

Five cards: d is displacement in m, v_i is initial velocity in m/s, v_f is final velocity in m/s, a is acceleration in m/s squared, and t is time in s. Caption: know any 3, find the other 2.

Every problem with constant acceleration involves the same five variables. If you know any three of them, the kinematic equations let you find the other two. Whenever a problem uses one of these quantities, write it down with its symbol and units before you do anything else.

The Four Equations

The four kinematic equations (constant acceleration only):

v_f = v_i + at (no d)
d = v_it + ½at² (no v_f)
v_f² = v_i² + 2ad (no t)
d = ½(v_f + v_i)t (no a)

Four kinematic equations in yellow boxes, each labeled with the variable it leaves out: v_f = v_i + at has no d. d = v_i t + one half a t squared has no v_f. v_f squared = v_i squared + 2ad has no t. d = one half (v_f + v_i) t has no a.

Each equation contains four of the five variables and leaves one out. That missing variable is the key to choosing which equation to use. The first one is the acceleration equation you already know, rearranged. The last one says displacement equals average velocity times time.

Choosing the Equation

List the 3 knowns
Name the unknown
Find the variable you don’t need
Use the equation without it

A table with two columns, Missing and Use. Missing d: use v_f = v_i + at. Missing v_f: use d = v_i t + one half a t squared. Missing t: use v_f squared = v_i squared + 2ad. Missing a: use d = one half (v_f + v_i) t.

To pick an equation, list the three quantities you know and the one you want. The fifth variable is the one you don’t know and don’t need. Choose the equation that leaves that variable out. For example, if a problem never mentions time and doesn’t ask for it, use v_f² = v_i² + 2ad.

Hidden Knowns

“From rest”: v_i = 0
“Comes to a stop”: v_f = 0
“Constant velocity”: a = 0

Two scenes. Top: a car at a starting line labeled starts from rest, v_i = 0. Bottom: a car stopped at a stop sign labeled comes to a stop, v_f = 0.

Problems often give you a known value in words instead of numbers. “Starts from rest” or “dropped” means the initial velocity is zero. “Comes to a stop” means the final velocity is zero. “Constant velocity” means the acceleration is zero. Read every problem carefully for these hidden knowns, because you usually need them to have three knowns.

Signs Matter

Pick a positive direction first
Slowing down: a has the opposite sign of v
Keep units on every number

A car moving to the right, which is labeled positive. A blue velocity arrow points right labeled v = +20 m/s, and a green acceleration arrow under the car points left, labeled a = −5.0 m/s squared. Caption: braking.

Velocity, displacement, and acceleration are vectors, so their signs matter. Before solving, choose which direction is positive. If an object is slowing down, its acceleration points opposite to its velocity, so it gets the opposite sign. A car moving at +20 m/s while braking has a negative acceleration, like −5.0 m/s².

Worked Examples

Try each problem on your own first, then check your work against the solution.

Example 1: From Rest

A car starts from rest and speeds up at 3.0 m/s² for 5.0 s. How far does it travel?

A car at a green starting line with a green arrow labeled a = 3.0 m/s squared and a label t = 5.0 s.

Knowns: v_i = 0 m/s (from rest), a = 3.0 m/s², t = 5.0 s. Unknown: d. Not needed: v_f
Equation: no v_f, so d = v_it + ½at²
Substitute: d = (0)(5.0 s) + ½(3.0 m/s²)(5.0 s)² = ½(3.0 m/s²)(25 s²)
Answer: d = 37.5 m, which rounds to 38 m (two significant figures)

Example 2: Stopping Distance

A car moving at 20.0 m/s brakes with an acceleration of −5.00 m/s² until it stops. How far does it travel while braking?

A car moving right toward a stop sign. A blue arrow labeled 20.0 m/s points right and a green arrow labeled −5.00 m/s squared points left.

Knowns: v_i = 20.0 m/s, v_f = 0 m/s (stops), a = −5.00 m/s². Unknown: d. Not needed: t
Equation: no t, so v_f² = v_i² + 2ad, rearranged to d = (v_f² − v_i²) / 2a
Substitute: d = (0 − (20.0 m/s)²) / (2 × −5.00 m/s²) = (−400. m²/s²) / (−10.0 m/s²)
Answer: d = 40.0 m

The two negative signs cancel, so the displacement is positive: the car keeps moving forward while it brakes.

Example 3: Landing Plane

A plane touches down at 70.0 m/s and slows to 10.0 m/s in 20.0 s. How far does it travel during this time?

A side view of a plane on a runway, with labels 70.0 m/s to 10.0 m/s and t = 20.0 s.

Knowns: v_i = 70.0 m/s, v_f = 10.0 m/s, t = 20.0 s. Unknown: d. Not needed: a
Equation: no a, so d = ½(v_f + v_i)t
Substitute: d = ½(10.0 m/s + 70.0 m/s)(20.0 s) = (40.0 m/s)(20.0 s)
Answer: d = 800. m

Example 4: Finding Time

A sled moving at 2.0 m/s speeds up at 0.80 m/s² down a hill until it reaches 6.0 m/s. How long does this take?

A sled on a snowy hill that slopes down to the right, with a green arrow along the slope labeled a = 0.80 m/s squared.

Knowns: v_i = 2.0 m/s, v_f = 6.0 m/s, a = 0.80 m/s². Unknown: t. Not needed: d
Equation: no d, so v_f = v_i + at, rearranged to t = (v_f − v_i) / a
Substitute: t = (6.0 m/s − 2.0 m/s) / 0.80 m/s² = 4.0 m/s / 0.80 m/s²
Answer: t = 5.0 s

Big Ideas

Constant a: five variables, four equations
Know 3 variables, find the rest
Use the equation missing the one you don’t need
Watch for hidden knowns and signs

Back to the essential question: when acceleration is constant, the four kinematic equations connect displacement, initial velocity, final velocity, acceleration, and time. If you know any three of these, you can choose the equation that leaves out the variable you don’t know and don’t need, and solve for the one you want. This wraps up our unit on linear motion.