Kinematics & Free Fall

You’re at a stoplight. The light turns green and you mash the gas. Your car accelerates at a steady rate. After 4 seconds, how fast are you going? How many car lengths have you covered? You don’t need calculus to answer either question — five equations handle every constant-acceleration problem you’ll ever see on the MCAT.

The trick isn’t memorizing all five. It’s knowing which one to grab. That decision takes one second once you know the rule, and it saves you minutes per problem.

The Big Five Kinematic Equations

These five equations describe motion with constant acceleration in one dimension. Each one connects four of the five kinematic variables and is missing one. The whole strategy: figure out which variable you don’t know and don’t need — then pick the equation that leaves it out.

How to Solve a Kinematics Problem

Same recipe every time:

1. Draw a quick diagram. Mark the start and the end.
2. Define a positive direction. Usually the direction of initial motion.
3. List knowns. Write down all five variables and circle the three you know.
4. Pick the equation that contains your three knowns plus the unknown you want.
5. Solve algebraically first, then plug numbers in.
6. Check units. If they don’t work out, the equation doesn’t either.

This routine adds maybe 15 seconds to a problem and saves 90 seconds in confused fumbling.

Worked Example 1

A car accelerates from rest at 4 m/s² for 5 seconds. How far does it travel?

• Knowns: $v_0 = 0$, $a = 4$ m/s², $t = 5$ s.
• Want: $\Delta x$.
• Missing (don’t need): $v$ (the final velocity).
• Pick: equation 2 (the one without $v$).

$\Delta x = v_0 t + \tfrac{1}{2}at^2 = 0(5) + \tfrac{1}{2}(4)(5^2) = \tfrac{1}{2}(4)(25) = 50 \text{ m}$

The car covers 50 m. (Sanity check: that’s about 10 car lengths in 5 seconds — feels right for a brisk acceleration.)

Worked Example 2

A motorcycle going 30 m/s slams on the brakes and stops in 60 m. What was its acceleration?

• Knowns: $v_0 = 30$ m/s, $v = 0$, $\Delta x = 60$ m.
• Want: $a$.
• Missing: $t$.
• Pick: equation 3.

$v^2 = v_0^2 + 2a(\Delta x) \Rightarrow 0 = 30^2 + 2a(60) \Rightarrow a = -\frac{900}{120} = -7.5 \text{ m/s}^2$

The negative sign means the acceleration points opposite to motion — i.e., it’s a deceleration. The brakes pulled the bike backward at 7.5 m/s² for as long as it took to stop.

Free Fall

An object in free fall is influenced only by gravity. No air resistance, no rocket thrust, no rope holding it up — just the steady downward pull of Earth.

In free fall, the only acceleration is $g$. Replace "$a$" with "$g$" (or "$-g$" if you define upward as positive) in the kinematic equations:

That’s it — the Big Five with $g$ swapped in for $a$.

Key Free-Fall Facts

These three facts solve a huge fraction of free-fall problems by themselves.

1. All objects fall at the same rate (no air resistance). A bowling ball and a tennis ball dropped together hit the ground simultaneously in a vacuum. Mass cancels: $F = mg$ → $a = mg/m = g$ for any mass.

2. An object thrown straight up decelerates at $g$ on the way up, momentarily stops at the peak ($v = 0$), then accelerates at $g$ on the way down. The speed at any given height is the same on the way up and the way down. Mirror-image trip.

3. Time up equals time down (when launched from and caught at the same height). Total air time is twice the time to reach the peak.

Useful Free-Fall Results

For an object dropped from rest ($v_0 = 0$) with $g \approx 10$ m/s²:

Notice the distances: 5, 20, 45, 80, 125. The pattern is $\tfrac{1}{2}gt^2$ — distance grows with the square of time. Doubling the fall time quadruples the distance fallen, not just doubles it. This is why falls from heights are so much more dangerous than they intuitively seem.