The four kinematic equations and when to use each

Every constant-acceleration problem in one dimension involves the same five quantities: displacement x, initial velocity v0, final velocity v, acceleration a and time t. The four kinematic equations each connect four of them and leave one out. Choosing an equation comes down to one question: which variable do you not know and not need? Use the equation that leaves that one out.

This is OpenStax College Physics 2e section 2.5, with t starting at 0 and x meaning the displacement (x minus x0).

The equations and what each leaves out

Some courses call these the SUVAT equations, after the letters s, u, v, a, t used for displacement, initial velocity, final velocity, acceleration and time. They are the same four equations.

All four assume the acceleration is constant. If it changes partway through, split the motion into stages and use the final velocity of one stage as the initial velocity of the next.

How to pick one

1. Draw the situation and choose a positive direction.
2. List the three quantities you know, with signs and units.
3. Name the one you want.
4. The fifth quantity is the one you neither know nor want. Pick the equation that leaves it out.
5. Solve, then ask whether the answer is reasonable.

That routine is the problem-solving strategy from OpenStax section 2.6, applied to kinematics. The examples below each hit a different equation.

Equation 1: no acceleration given

A car speeds up uniformly from 10.0 m/s to 30.0 m/s in 5.00 s. How far does it go?

Known: v0, v, t. Wanted: x. Not given and not needed: a. Use equation 1:

x = (10.0 + 30.0) / 2 x 5.00 = 100 m

(If you did want the acceleration, equation 2 gives (30.0 - 10.0) / 5.00 = 4.00 m/s^2.)

Equation 2: no displacement involved

A cart moving at 10.0 m/s accelerates at 2.00 m/s^2 for 5.00 s. What is its final speed?

Known: v0, a, t. Wanted: v. Displacement plays no part, so use equation 2:

v = 10.0 + 2.00 x 5.00 = 20.0 m/s

Equation 3: no final velocity involved

A car starts from rest and accelerates at 3.00 m/s^2 for 4.00 s. How far does it travel?

Known: v0 = 0, a, t. Wanted: x. Final velocity is not needed, so use equation 3:

x = 0 + (1/2)(3.00)(4.00)^2 = 24.0 m

The most common slip here is dropping the 1/2, which gives 48.0 m.

Equation 4: no time given

A car moving at 20.0 m/s brakes at 5.00 m/s^2 and stops. How far does it travel while braking?

Take the direction of motion as positive, so a = -5.00 m/s^2 and v = 0. Time is not given and not asked for, so use equation 4:

0 = 20.0^2 + 2(-5.00) x, so x = 400 / 10.0 = 40.0 m

Equation 4 also explains a fact about braking: stopping distance goes as the square of the starting speed. At 40.0 m/s, with the same deceleration, the car needs 160 m, four times as far.

A second example: a plane starts from rest, accelerates at 2.00 m/s^2 and needs 60.0 m/s to take off. The runway must be at least 60.0^2 / (2 x 2.00) = 900 m long.

Free fall is the same equations

An object in free fall has a = -9.80 m/s^2 if you take up as positive. Nothing else changes.

• A ball thrown straight up at 14.7 m/s rises until v = 0. Equation 4: 0 = 14.7^2 - 2(9.80) y, so y = 11.0 m. Equation 2: 0 = 14.7 - 9.80 t, so it takes 1.50 s to reach the top.
• A ball thrown up at 10.0 m/s has velocity v = 10.0 - 9.80 x 1.50 = -4.70 m/s after 1.50 s. The negative sign means it is already on the way down.
• A stone dropped from 20.0 m: y = (1/2)(9.80) t^2 with y = 20.0 m (taking down as positive here), so t = sqrt(2 x 20.0 / 9.80) = 2.02 s.

At the top of a throw the velocity is zero but the acceleration is still 9.80 m/s^2 downward. That is one of the most tested ideas in the unit.

When there are two answers

Equation 4 gives v^2, not v, so the square root has two signs. Sometimes both are real answers.

A ball is thrown straight up at 10.0 m/s. When is it 3.00 m above the launch point? Take up as positive, so a = -9.80 m/s^2 and x = 3.00 m.

• Equation 4: v^2 = 10.0^2 - 2(9.80)(3.00) = 41.2, so v = +6.42 m/s or -6.42 m/s
• Equation 2 for each: t = (6.42 - 10.0) / -9.80 = 0.365 s (going up) and t = (-6.42 - 10.0) / -9.80 = 1.68 s (coming down)

Both times are positive, so both count. The ball passes 3.00 m on the way up and again on the way down, at the same speed in opposite directions. The two times add to 2.04 s, which is exactly the time it takes to return to the launch point, 2 x 10.0 / 9.80.

If you ask when it reaches 6.00 m, v^2 comes out negative. That is not an error to work around. It means the ball never gets that high: its peak is 10.0^2 / (2 x 9.80) = 5.10 m.

When a root gives a negative time, throw it out. It describes a moment before the motion started.

Mistakes to watch

• Mixing sign conventions. If up is positive, a = -9.80 m/s^2 and a falling object has a negative velocity. Decide once per problem.
• Using the equations when acceleration changes. Split the motion into stages.
• Treating "decelerating" as "negative acceleration." Deceleration means acceleration opposite to the velocity. A car moving in the negative direction and slowing down has a positive acceleration.
• Forgetting the square root in equation 4. Leaving it out gives v^2, and the units come out as m^2/s^2, which is a quick way to catch it.

Practice

The kinematics calculator takes any three of the five quantities, solves for the other two and tells you which equation it used, including both roots when there are two. Once one dimension feels automatic, move on to projectile motion equations with worked examples, which runs these equations in two directions at once, with the projectile motion calculator to check your work. How to study for a physics exam covers how to practice problems like these so they hold up under time pressure. The same knowns-and-unknowns table works in chemistry, where stoichiometry step by step is a chain of conversions rather than a choice of equation.

Kinematics is units 2 and 3 of Encodr's free College Physics I deck.