3.1 Kinematics & Motion

Kinematics — the study of motion, independent of its causes — is the mathematical backbone of every mechanics question on the physics portion of the Navy Advanced Programs Test (NAPT). Free-fall drops, thrown objects, accelerating vehicles, and word problems asking how far, how fast, or how long an object travels all reduce to the same handful of relationships covered in this section. Learn the four kinematics equations cold, and learn to split any projectile problem into a horizontal piece and a vertical piece, and a large share of the mechanics questions on the NAPT become straightforward arithmetic.

Distance, Displacement, Speed, and Velocity

Not every question asking how far or how fast means the same thing in physics, and the exam relies on you knowing the difference.

• Distance is a scalar quantity — total path length traveled, with no direction attached. Walk 3 meters east and then 3 meters west, and you have traveled a distance of 6 meters.
• Displacement is a vector quantity — the straight-line change in position, including direction. In that same example, your displacement is 0 meters because you ended up back where you started.
• Speed is distance divided by time (a scalar). Velocity is displacement divided by time (a vector) — it carries both magnitude and direction.
• Acceleration is the rate at which velocity changes over time, measured in meters per second squared (m/s²) in the metric system or feet per second squared (ft/s²) in the imperial system. An object speeding up, slowing down, or changing direction is all accelerating — deceleration is simply acceleration directed opposite to the motion.

The Four Kinematics Equations

For motion with constant acceleration — the case tested almost exclusively on the NAPT — four equations describe everything: where an object is, how fast it is moving, and how long it takes to get there. Each equation omits one of the five key variables (v0, v, a, t, Δx), so picking the right equation just means picking the one that does not require the variable you were not given.

Where v0 = initial velocity, v = final velocity, a = acceleration, t = time, and Δx = displacement.

Worked Example 1: An Accelerating Boat

A small patrol boat starts from rest alongside the pier and accelerates at a constant 2.5 m/s² for 6 seconds before leveling off.

Find the final velocity. Since we know v0 (0 m/s), a (2.5 m/s²), and t (6 s), use v = v0 + at:

v = 0 + (2.5)(6) = 15 m/s

Find the distance covered. Use Δx = v0t + ½at²:

Δx = (0)(6) + ½(2.5)(6²) = 0 + ½(2.5)(36) = 45 meters

Because acceleration is constant, velocity grows in a straight line over time — the chart below plots the boat's velocity at each second of the run.

Worked Example 2: Free Fall

A wrench slips from a sailor's hand and falls from a height of 44.1 meters above the deck. Ignoring air resistance, gravity accelerates it downward at g = 9.8 m/s².

Find the time to hit the deck. Starting from rest, use Δx = v0t + ½at² with v0 = 0:

44.1 = ½(9.8)t² → t² = 44.1 / 4.9 = 9 → t = 3 seconds

Find the impact velocity. Use v = v0 + at:

v = 0 + (9.8)(3) = 29.4 m/s

Free-fall problems are just kinematics problems with a = g = 9.8 m/s² (or 32.2 ft/s² in imperial units) substituted in — nothing new to memorize.

Projectile Motion: Two Independent Problems in One

A projectile is any object moving under gravity alone after being launched, thrown, or dropped. The single most important idea for the NAPT is that projectile motion is really two separate, independent motions happening at the same time:

Because the two directions never affect each other, you solve the vertical problem first to find time of flight, then plug that same time into the horizontal problem to find range (horizontal distance traveled).

Worked Example 3: Horizontal Launch

A signal flare is launched horizontally at 20 m/s from a platform 78.4 meters above the water.

Step 1 — Vertical: find time of flight. The flare has no initial vertical velocity, so use Δx = ½gt²:

78.4 = ½(9.8)t² → t² = 78.4 / 4.9 = 16 → t = 4 seconds

Step 2 — Horizontal: find range. Horizontal velocity never changes, so distance = velocity × time:

range = (20 m/s)(4 s) = 80 meters

Bonus — impact speed. The vertical velocity at splashdown is v = gt = (9.8)(4) = 39.2 m/s. Combining the horizontal (20 m/s) and vertical (39.2 m/s) components with the Pythagorean theorem gives a total impact speed of about 44 m/s — noticeably faster than either component alone, because velocity is a vector.

This same two-step method — solve the vertical component for time, then use that time horizontally — handles every horizontal-launch projectile problem you will see, whether it is a dropped tool, a thrown heaving line, or a launched flare.

Section Takeaways

• Four kinematics equations cover any constant-acceleration motion; pick the one that omits the variable you were not given.
• Distance and speed are scalars (no direction); displacement and velocity are vectors (direction matters).
• Free fall is kinematics with a = g = 9.8 m/s² (32.2 ft/s²) substituted in.
• Projectile motion splits into two independent problems: horizontal (constant velocity) and vertical (constant acceleration g).
• Solve the vertical component for time of flight first, then use that time to find horizontal range.