The four kinematics equations
When acceleration is constant, four linked equations let you skip the calculus and connect any three of x, v, a, and t.
The four equations
Each equation deliberately leaves one variable out, and that is how you choose which to use. No x? Use v = v0 + at. No t? Use v^2 = v0^2 + 2ax. No v? Use x = x0 + v0 t + (1/2)at^2. No a? Use x = (1/2)(v0 + v)t. All four assume constant acceleration and straight-line motion.
v = v_0 + atChoosing the right one
List what the problem gives and what it asks for, then pick the equation whose missing variable matches the one you neither know nor need. Watch signs: a negative acceleration means the acceleration points along the negative axis, and a negative displacement means the object ended behind the start. Free fall is the most common case, with a = -9.8 m/s^2 taking upward as positive.
v^2 = v_0^2 + 2a\Delta xWorked example
A car traveling 12 m/s brakes with constant acceleration of -3 m/s^2. To find stopping time, use v = v0 + at: 0 = 12 - 3t gives t = 4 s. For stopping distance, use v^2 = v0^2 + 2ax since time is now extra information: 0 = 144 + 2(-3)x gives x = 24 m. Check with x = v0 t + (1/2)at^2 = 48 - 24 = 24 m.
x = \frac{v^2 - v_0^2}{2a} = \frac{0 - 144}{-6} = 24 \text{ m}Flashcard
Velocity under constant acceleration, no position
Check your understanding
A train slows from 30 m/s to 10 m/s while covering 200 m of track. What is its acceleration, assumed constant?