Kinematic Equations for Constant Acceleration

The kinematic equations (SUVAT equations) describe motion in a straight line with constant acceleration, linking displacement s, initial velocity u, final velocity v, acceleration a and time t: v = u + at, s = ut + ½at², v² = u² + 2as, and s = ½(u + v)t. In brief: each equation has four of the five variables, so if you know three, you can find the other two. They hold only if the acceleration is constant, and the signs of the vectors must be chosen consistently.

These equations are the main tools for solving problems on cars, falling objects and projectiles. This guide lists the equations and shows where they come from, gives a method for choosing the right one, compares the equations in a table, works through three problems and corrects common mistakes.

What are the kinematic equations?

The symbols are: s displacement (m), u initial velocity (m/s), v final velocity (m/s), a constant acceleration (m/s²), t time (s). The four equations are:

1. v = u + at
2. s = ut + ½at²
3. v² = u² + 2as
4. s = ½(u + v)t

In some books, x replaces s, v₀ or v_i replaces u, and v_f replaces v. The equations work in one dimension, and for motion in a plane they are applied separately to the horizontal and vertical components.

Where do the equations come from?

• v = u + at is the definition of acceleration, a = (v − u) ÷ t, rearranged.
• s = ½(u + v)t: with constant acceleration, the average velocity is the mean of u and v, and displacement = average velocity × time. It is also the area of the trapezoid on the velocity-time graph.
• s = ut + ½at² comes from substituting v = u + at into the previous equation: s = ½(u + u + at)t = ut + ½at².
• v² = u² + 2as comes from eliminating t between equations 1 and 4: t = (v − u) ÷ a, so s = ½(u + v)(v − u) ÷ a = (v² − u²) ÷ 2a.

How do you choose an equation?

1. Draw a sketch and choose a positive direction. Write the data with signs.
2. List the five variables and mark the three that are known and the one that is wanted.
3. Choose the equation which contains the wanted variable and the three known ones, but not the fifth.
4. Solve algebraically, then substitute, and check the unit and the size of the answer.

The equations compared

Worked example: three problems

1. Accelerating car. A car at 10 m/s accelerates at 2.0 m/s² for 5.0 s. Find the final velocity and the distance.
   • v = u + at = 10 + 2.0 × 5.0 = 20 m/s.
   • s = ut + ½at² = 10 × 5.0 + ½ × 2.0 × 5.0² = 50 + 25 = 75 m.
   • Check with s = ½(u + v)t = ½ × (10 + 20) × 5.0 = 75 m.

2. Braking distance. A car at 20 m/s brakes with a constant deceleration of 5.0 m/s² until it stops. Take the direction of motion as positive, so a = −5.0 m/s² and v = 0.
   • v² = u² + 2as gives 0 = 400 + 2 × (−5.0) × s, so s = 400 ÷ 10 = 40 m.
   • Time: t = (v − u) ÷ a = (0 − 20) ÷ (−5.0) = 4.0 s.
   • If the speed were 30 m/s, the stopping distance would be 900 ÷ 10 = 90 m, which is 2.25 times as much, because the distance is proportional to the square of the speed.

3. Acceleration from distance. A cyclist starts from rest and travels 36 m in 6.0 s with constant acceleration. u = 0, s = 36 m, t = 6.0 s.
   • s = ½at² gives a = 2s ÷ t² = 72 ÷ 36 = 2.0 m/s².
   • Final velocity v = at = 2.0 × 6.0 = 12 m/s.

How were the equations found?

Medieval scholars at Merton College, Oxford, and Nicole Oresme in Paris described uniformly accelerated motion in the 1300s. Galileo tested the law s ∝ t² experimentally, using balls on inclined planes, and published it in Two New Sciences (1638). The letters s, u, v, a, t became the standard notation in British school courses in the twentieth century.

Which mistakes are common?

Mistake 1: Using the equations when acceleration is not constant

They do not apply to a car with changing acceleration, or to a falling body with strong air resistance.

Mistake 2: Inconsistent signs

If up is positive, then gravity is a = −9.8 m/s². Do not mix up and down.

Mistake 3: Forgetting that u = 0 and v = 0 are data

“Starts from rest” gives u = 0, and “comes to a stop” gives v = 0. These are the third known variable.

Mistake 4: Taking the square root without considering both signs

When solving for v from v², choose the root that agrees with the direction of the motion.