Scalars & Vectors

Your car’s odometer reads 150 miles after a road trip. Your GPS says you ended up only 80 miles from where you started. Both numbers are correct — they just measure different things. The odometer counted every mile your tires turned, even the loops, U-turns, and detours. The GPS only cares about your straight-line distance and direction from start to end.

The odometer is a scalar — a pure number, no direction. The GPS is a vector — number plus direction.

This distinction looks small, but it’s the foundation of every motion problem on the MCAT. Get it wrong and you’ll mix up speed and velocity, distance and displacement, mass and weight — all the classic trap-question pairs.

Scalars

A scalar is fully described by a number and a unit — nothing else needed.

Common MCAT scalars:

• Mass (5 kg)
• Temperature (310 K)
• Speed (20 m/s)
• Distance (150 m)
• Energy (500 J)
• Time (3 s)
• Pressure (101,325 Pa)

You can add, subtract, multiply, and divide scalars with ordinary arithmetic. 5 kg + 3 kg = 8 kg, no questions asked.

Vectors

A vector has both magnitude and direction. You need both to fully describe it. “50 newtons” doesn’t mean anything by itself — 50 newtons in which direction?

Common MCAT vectors:

• Displacement (80 m east)
• Velocity (20 m/s north)
• Acceleration (9.8 m/s² downward)
• Force (50 N at 30° above horizontal)
• Momentum (15 kg·m/s to the right)

On paper, vectors are arrows. The arrow’s length shows magnitude; the arrow’s direction shows where it’s pointing. A long arrow = strong force or fast velocity; a short arrow = weak force or slow velocity.

Vector Components

Any vector pointing somewhere “diagonal” can be split into two perpendicular pieces — one along the x-axis, one along the y-axis. These are called components, and breaking a vector apart this way is called resolving it.

Why bother? Because the math is much easier when everything points along an axis. Splitting a 50 N diagonal force into “30 N to the right” and “40 N up” turns one tricky 2D problem into two easy 1D problems.

If a vector $\vec{A}$ has magnitude $A$ and makes an angle $\theta$ with the x-axis:

To go backward — from components to the full vector — use the Pythagorean theorem and basic trig:

Key Angles to Know

You won’t have a calculator on the MCAT. Memorize these trig values cold:

Worked Example

A passenger plane is climbing at 200 m/s along a flight path 30° above the horizontal. What are the horizontal and vertical components of its velocity?

• $v_x = v\cos\theta = 200 \cos(30°) = 200 \times 0.87 \approx 173$ m/s (forward).
• $v_y = v\sin\theta = 200 \sin(30°) = 200 \times 0.5 = 100$ m/s (upward).

So the plane covers ground at 173 m/s while climbing at 100 m/s. Sanity check: $\sqrt{173^2 + 100^2} \approx \sqrt{30{,}000 + 10{,}000} = \sqrt{40{,}000} = 200$ m/s. ✓

Negative Components and Sign Conventions

A component can be negative. The convention almost everyone uses (and the MCAT defaults to):

• Right is positive x; left is negative x.
• Up is positive y; down is negative y.

So a vector pointing “30° south of west” has a negative x-component (leftward) and a negative y-component (downward).

Always pin down your coordinate system before solving the problem. The standard “x right, y up” works for most setups, but inclined-plane problems usually tilt the axes to lie along the slope (more on that in §1.11). Whatever you choose, stick with it for the entire problem.