🚀The Kinematic Equations
The Big Four
When acceleration is constant, four equations relate position, velocity, acceleration, and time. These are the most-used equations in all of AP Physics 1.
Each equation uses a different combination of variables. Your strategy is: identify what you know, identify what you want, then pick the equation that connects them.
Variables: x (displacement), v₀ (initial velocity), v (final velocity), a (acceleration), t (time)
📐 The Four Kinematic Equations
Valid only when acceleration is constant:
v = v_0 + atx = v_0 t + (1)/(2)at²v² = v_0² + 2axx = (v_0 + v)/(2) · tThink About It
Which kinematic equation would you use if you know initial velocity, final velocity, and displacement — but NOT time?
Strategy for Kinematic Problems
Follow these steps every time: 1. Draw a diagram and define positive direction 2. List knowns: write down every given quantity with units 3. Identify unknown: what are you solving for? 4. Choose equation: pick the equation that has your knowns and unknown 5. Solve algebraically first, then substitute numbers 6. Check units and signs: does the answer make physical sense?
Common mistake: forgetting that "starts from rest" means v₀ = 0, and "comes to a stop" means v = 0.
✏️ Worked Example 1
Problem: A ball is dropped from a 45 m tall building. How long does it take to hit the ground? (g = 10 m/s²)
✏️ Worked Example 2
Problem: A car traveling at 20 m/s brakes and stops in 40 m. What is the acceleration?
⚠️ Common Mistakes
Misconception: You can use the kinematic equations even when acceleration is changing.
✓ Correct thinking: The four kinematic equations are valid ONLY when acceleration is constant throughout the interval. If acceleration varies, you need calculus or energy methods.
Why: Every kinematic equation is derived by integrating constant acceleration. They break down the moment acceleration changes.
Misconception: "Starts from rest" doesn't have to be written down — you can remember it.
✓ Correct thinking: Always write v₀ = 0 explicitly when an object starts from rest, and v = 0 when it comes to a stop. Omitting this is one of the leading causes of sign errors.
Why: In multi-step problems under time pressure, skipping written knowns leads to using the wrong equation or plugging in the wrong value.
Misconception: Solving for time using v² = v₀² + 2ax gives the correct answer.
✓ Correct thinking: That equation has no t in it — you cannot solve for time from it directly. Choose x = v₀t + ½at² or v = v₀ + at when time is the unknown.
Why: Picking the wrong kinematic equation is the most common procedural error on free-response questions.
📝 Practice Problems
Try these problems. Check your answer when ready.
A train starts from rest and accelerates at 2 m/s². How fast is it going after 10 seconds?
v = v_0 + atA rock is dropped from a bridge. It hits the water 4 seconds later. How high is the bridge? (g = 10 m/s²)
x = v_0 t + (1)/(2)at²A car going 30 m/s brakes to a stop in 5 seconds. How far does it travel while braking?
A rocket accelerates from rest to 400 m/s over a distance of 800 m. What is its acceleration, and how long did the burn last?
Two cars start from the same position. Car A moves at constant 20 m/s. Car B starts from rest and accelerates at 4 m/s². At what time does Car B catch up to Car A? How far have they traveled?
x_A = x_B ⇒ 20t = (1)/(2)(4)t²A ball is thrown upward at 25 m/s from the edge of a 45 m tall cliff. How long until it hits the ground at the base of the cliff? (g = 10 m/s², take up as positive, ground is at y = −45 m)
y = v_0 t - (1)/(2)gt²A police car at rest begins chasing a speeder traveling at a constant 30 m/s. The police car accelerates at 5 m/s². How long before the police car reaches the same speed as the speeder? How far behind is the police car at that moment?
Finished reading through this lesson?