The formula
- Velocity from time
v = v₀ + at- Position from time
x = x₀ + v₀t + ½at²- Velocity from displacement
v² = v₀² + 2a(x - x₀)- Position from average velocity
x = x₀ + ½(v₀ + v)t
What the symbols mean
| Symbol | Meaning | Unit |
|---|---|---|
v | Final velocity | m/s |
v₀ | Initial velocity | m/s |
x | Final position | m |
x₀ | Initial position | m |
a | Constant acceleration | m/s² |
t | Elapsed time | s |
When it applies
- Acceleration is constant across the whole interval. An acceleration that changes partway through breaks every one of these equations.
- The motion is along one straight line, or you have already split a two-dimensional problem into independent horizontal and vertical components.
- You know three of the five quantities and want a fourth. Choose the equation that omits the one you neither know nor need.
Worked example
Problem. A car traveling at 12 m/s brakes with a constant acceleration of 3.0 m/s² opposite its motion. How far does it travel before stopping, and how long does that take?
- List the knowns with the direction of travel as positive: v₀ = 12 m/s, v = 0 m/s, a = -3.0 m/s². Distance and time are both unknown, so start with the equation that leaves time out.
- Use v² = v₀² + 2a(x - x₀): 0 = (12 m/s)² + 2(-3.0 m/s²)(x - x₀).
- Solve for the displacement: x - x₀ = 144 / 6.0 = 24 m.
- Now use v = v₀ + at for the time: 0 = 12 + (-3.0)t, so t = 12 / 3.0 = 4.0 s.
Answer. The car covers 24 m and takes 4.0 s to stop.
Common mistakes
- Applying the equations across an interval where the acceleration changes, such as a car that brakes and then coasts. Split the motion into constant-acceleration segments instead.
- Dropping the sign on acceleration. Slowing down is a negative acceleration only if you chose the direction of motion as positive, and that choice has to stay consistent across all four equations.
- Mixing a horizontal quantity into a vertical equation. In two dimensions the components are independent and share only the time.
- Setting v₀ to zero out of habit. It is zero only when the object genuinely starts at rest.
Related formulas
- Work-energy theorem:
W_net = ΔKE = ½mv² - ½mv₀² - Newton's second law:
ΣF = ma - Projectile motion formulas:
v₀ₓ = v₀·cos θ₀ and v₀ᵧ = v₀·sin θ₀