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AP Physics 1: 1D Motion & The “Big 4” Equations

Exam Tip: AP Physics 1 is strictly about Constant Acceleration. If acceleration changes (like a car jerking), these equations DO NOT work.

1. The 5 Variables of Motion

A grid showing the symbols for displacement, initial velocity, final velocity, acceleration, and time. — The “High Five” variables of Kinematics. Every problem gives you 3 and asks for 1.

Every kinematics problem involves exactly 5 variables. Your first step is always to list them:

$\Delta x$ : Displacement (m)
$t$ : Time interval (s)
$v_0$ : Initial Velocity (m/s)
$v$ : Final Velocity (m/s)
$a$ : Acceleration (m/s²)

2. The “Big 4” Equations

These are provided on your AP Formula Sheet. The secret to speed is knowing which variable is missing.

Equation	Missing Variable?	Best Use Case
$v = v_0 + at$	Displacement ( $\Delta x$ )	Finding velocity or time.
$\Delta x = v_0t + \frac{1}{2}at^2$	Final Velocity ( $v$ )	Dropping objects or finding distance.
$v^2 = v_0^2 + 2a\Delta x$	Time ( $t$ )	Stopping distance problems.
$\Delta x = \frac{1}{2}(v_0 + v)t$	Acceleration ( $a$ )	Average velocity problems.

3. Vertical Motion (Free Fall)

Side-by-side comparison of a dropped ball versus a ball thrown upward, showing velocity and acceleration vectors. — Left: Dropping a ball ( $v_0 = 0$ ). Right: Throwing a ball up. Notice that acceleration ( $g$ ) is always pointing down, even when the ball goes up!

When an object is in the air (thrown up or dropped), only gravity acts on it.

Dropping an Object

$v_0 = 0$
$a = -9.8 \, \text{m/s}^2$ (or $-10$ on MCQ)
Displacement $\Delta y$ is negative.

Throwing Upward

Velocity at peak = $0 \, \text{m/s}$
Acceleration is always $-9.8$ , even at the top!
Time up = Time down (if landing on same level).

4. AP-Style Practice Question

Question: A car traveling at $20 \, \text{m/s}$ applies the brakes and stops over a distance of $50 \, \text{m}$ . What was the car’s acceleration?

▶ Click to see Solution

Step 1: List Variables
$v_0 = 20$ , $v = 0$ (stopped), $\Delta x = 50$ , $a = ?$
Time ( $t$ ) is missing!

Step 2: Choose Equation
Use the “Time Independent” equation: $v^2 = v_0^2 + 2a\Delta x$

Step 3: Solve
$0^2 = (20)^2 + 2(a)(50)$
$0 = 400 + 100a$
$-400 = 100a$
$a = -4 \, \text{m/s}^2$

5. Level Up: Harder Practice Problems

Question 2 (Free Fall): A student throws a ball straight up into the air with an initial speed of $30 \, \text{m/s}$ . How long does it take for the ball to reach its maximum height? (Assume $g = 10 \, \text{m/s}^2$ )

▶ Click for Solution

Key Concept: At the maximum height, the velocity is zero ( $v = 0$ ).

Variables: $v_0 = 30$ , $v = 0$ , $a = -10$ (gravity points down!), $t = ?$

Equation: $v = v_0 + at$

Solve:
$0 = 30 + (-10)t$
$-30 = -10t$
$t = 3.0 \, \text{s}$

Question 3 (Symbolic Derivation): A sprinter starts from rest and accelerates at a constant rate $a$ for a time $t$ . Derive an expression for the final velocity $v_f$ in terms of the distance traveled $D$ and the acceleration $a$ .

▶ Click for Solution

Step 1: Identify Knowns
Initial velocity $v_0 = 0$ . Acceleration = $a$ . Distance $\Delta x = D$ .
Note: We need to eliminate time ( $t$ ) because it is not in the requested answer!

Step 2: Choose Equation
Use the time-independent equation: $v^2 = v_0^2 + 2a\Delta x$

Step 3: Solve
$v_f^2 = (0)^2 + 2aD$
$v_f = \sqrt{2aD}$