Translational Motion

Translational motion is the study of objects moving through space — how fast, in what direction, and how that motion changes over time. The MCAT tests these ideas in biological contexts: blood flow, nerve impulse timing, or a projectile modeled after a thrown ball. Solid intuition here makes every later mechanics topic easier.

Priority labels: Must know = cold; Know the logic = mechanism not names; Passage-level = recognize, don't memorize; Optional = skippable.

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Units and Dimensions

The SI System and Why It Matters

Must know

The MCAT uses SI units almost exclusively. The three base units for mechanics are meter (m, length), kilogram (kg, mass), and second (s, time). Every other mechanical quantity — velocity, acceleration, force, energy — is a derived unit built from these (e.g. velocity = m/s, force = $\text{kg}\cdot\text{m/s}^2$).

Dimensional Analysis

Must know

Dimensional analysis tracks units through a calculation like algebra. If the units on both sides of an equation don't match, the equation is wrong — a reliable error-check. You can only add or subtract quantities with the same dimensions, but can multiply or divide different dimensions to create new ones (e.g. velocity $[L][T]^{-1}$, acceleration $[L][T]^{-2}$, force $[M][L][T]^{-2}$).

Common Prefixes

Must know

Memorize these prefixes cold.

Prefix	Symbol	Factor
nano-	n	$10^{-9}$
micro-	μ	$10^{-6}$
milli-	m	$10^{-3}$
centi-	c	$10^{-2}$
kilo-	k	$10^{3}$

In biology, distances are often given in cm or mm — convert to meters before plugging into kinematic equations. Forgetting this is a common arithmetic mistake.

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Quick check: A nerve impulse travels at 70 m/s. How fast is this in km/s?

Answer: $70 \text{ m/s} \times \frac{1 \text{ km}}{1000 \text{ m}} = 0.070 \text{ km/s}$. ✓

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Vectors and Components

Scalars vs. Vectors — The Core Distinction

Must know

• Scalars: magnitude alone — speed, distance, mass, temperature, time, energy.
• Vectors: magnitude and direction — velocity, displacement, acceleration, force, momentum.

Vectors are written with an arrow ($\vec{v}$); the magnitude is $|\vec{v}|$ or $v$.

Resolving a Vector into Components

Must know

Any 2D vector breaks into a horizontal and vertical component — this converts a 2D problem into two independent 1D problems, the single most useful skill in MCAT mechanics. For a vector $\vec{A}$ at angle $\theta$ from the positive x-axis:

$A_x = A \cos\theta \qquad A_y = A \sin\theta$

$A = \sqrt{A_x^2 + A_y^2} \qquad \theta = \arctan\!\left(\frac{A_y}{A_x}\right)$

By convention $\theta$ is measured counterclockwise from the +x-axis. Assign the correct sign to $A_y$ (positive up, negative down).

Worked Example: Decomposing a Velocity Vector

Must know

A soccer ball is kicked at $20 \text{ m/s}$ at $30°$ above the horizontal.

$v_x = 20 \cos 30° \approx 17.3 \text{ m/s} \qquad v_y = 20 \sin 30° = 10.0 \text{ m/s}$

Sanity check: $v_x > v_y$ makes sense — $30°$ is shallow, so most velocity is horizontal. ✓

Must know these exact trig values:

Angle	$\cos\theta$	$\sin\theta$
$0°$	1	0
$30°$	$\approx 0.87$	$0.50$
$45°$	$\approx 0.71$	$\approx 0.71$
$60°$	$0.50$	$\approx 0.87$
$90°$	0	1

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Quick check: A force vector has components $F_x = 3 \text{ N}$ and $F_y = 4 \text{ N}$. What is its magnitude?

Answer: $F = \sqrt{3^2 + 4^2} = 5 \text{ N}$. (The 3-4-5 triangle.) ✓

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Vector Addition and Subtraction

Graphical: Head-to-Tail

Must know

Place the tail of the second vector at the head of the first; the resultant $\vec{R} = \vec{A} + \vec{B}$ runs from the tail of $\vec{A}$ to the head of $\vec{B}$. Addition is commutative: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$.

By Components — The Reliable Method

Must know

1. Resolve each vector into $x$ and $y$ components.
2. Add components separately: $R_x = A_x + B_x$, $R_y = A_y + B_y$.
3. Magnitude $R = \sqrt{R_x^2 + R_y^2}$; direction $\theta = \arctan(R_y / R_x)$.

Vector Subtraction

Must know

Subtracting = adding the negative: $\vec{A} - \vec{B} = \vec{A} + (-\vec{B})$, where $-\vec{B}$ has the same magnitude but opposite direction. This is key for change in velocity ($\Delta \vec{v} = \vec{v}_f - \vec{v}_i$). Do not subtract magnitudes if the directions differ — use components.

Worked Example: Adding Displacement Vectors

Must know

A red blood cell travels $3 \text{ cm}$ east, then $4 \text{ cm}$ north.

$R = \sqrt{3^2 + 4^2} = 5 \text{ cm}, \qquad \theta = \arctan\!\left(\tfrac{4}{3}\right) \approx 53° \text{ N of E}$

Sanity check: displacement (5 cm) < total distance (7 cm). Displacement is always ≤ distance. ✓

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Quick check: A swimmer swims $5 \text{ m}$ east, then $5 \text{ m}$ west. What is her displacement? What distance did she travel?

Answer: Displacement = $0 \text{ m}$ (back to start). Distance = $10 \text{ m}$.

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Speed, Velocity (Average and Instantaneous)

Distance vs. Displacement

Must know

• Distance ($d$): total path length — always positive, scalar.
• Displacement ($\Delta \vec{r} = \vec{r}_f - \vec{r}_i$): straight-line change in position, with direction — vector.

These are equal only for straight-line motion with no backtracking.

Average Speed vs. Average Velocity

Must know

$\bar{v}_{\text{speed}} = \frac{\text{total distance}}{t} \qquad \bar{\vec{v}} = \frac{\Delta \vec{r}}{\Delta t}$

The MCAT loves scenarios where these differ: running a full 400 m lap in 80 s gives average speed $5 \text{ m/s}$ but average velocity $= 0$ (zero displacement).

Instantaneous Velocity

Know the logic

Instantaneous velocity is the velocity at one moment — formally the derivative of position, $\vec{v} = d\vec{r}/dt$. You don't take derivatives on the MCAT; instead, on a position-vs-time graph the instantaneous velocity is the slope of the tangent line. Steeper slope = greater speed; negative slope = motion in the −direction; horizontal line = at rest.

Worked Example: Average Speed vs. Average Velocity

Must know

A red blood cell travels 12 cm through a looping capillary in 0.60 s, ending 5 cm (straight-line) from start.

$\bar{v}_{\text{speed}} = \frac{12}{0.60} = 20 \text{ cm/s} \qquad |\bar{\vec{v}}| = \frac{5}{0.60} \approx 8.3 \text{ cm/s}$

Sanity check: average speed > average velocity magnitude, as expected for a curved path. ✓

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Quick check: On a position-vs-time graph, the slope from $t = 0$ to $4 \text{ s}$ is $+6 \text{ m/s}$; from $t = 4$ to $8 \text{ s}$ the graph is horizontal. What is the average velocity over the full 8 s?

Answer: Displacement $= 6 \times 4 = 24 \text{ m}$ (then 0). Average velocity $= 24/8 = 3 \text{ m/s}$.

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Acceleration

Building the Concept

Must know

Acceleration is how velocity changes over time. Because velocity is a vector, acceleration is nonzero whenever the magnitude or direction of velocity changes. A car at constant speed around a curve is accelerating because its direction changes — important for circular motion later.

Average and Instantaneous Acceleration

Must know

$\bar{\vec{a}} = \frac{\Delta \vec{v}}{\Delta t} = \frac{\vec{v}_f - \vec{v}_i}{\Delta t} \quad (\text{units } \text{m/s}^2)$

Instantaneous acceleration is the slope of the tangent on a velocity-vs-time graph.

Reading Kinematic Graphs

Must know

These graph relationships:

Graph	Slope	Area
Position vs. time	velocity	—
Velocity vs. time	acceleration	displacement
Acceleration vs. time	—	change in velocity

A curved $x$-vs-$t$ graph means changing velocity (nonzero acceleration). A straight $v$-vs-$t$ graph means constant acceleration.

Uniformly Accelerated Motion and the Kinematic Equations

Must know

When acceleration is constant, four kinematic equations describe 1D motion:

v_f = v_i + at \tag{1}

\Delta x = v_i t + \frac{1}{2}at^2 \tag{2}

v_f^2 = v_i^2 + 2a\Delta x \tag{3}

\Delta x = \frac{v_i + v_f}{2} \cdot t \tag{4}

Strategy: list the five variables ($v_i$, $v_f$, $a$, $\Delta x$, $t$), identify the three you know plus your unknown, and pick the equation missing the unused variable. Equation (3) is useful when time is neither given nor asked.

Free Fall and Gravitational Acceleration

Must know

Free fall is motion under gravity alone (no air resistance), with $g = 9.8 \text{ m/s}^2 \approx 10 \text{ m/s}^2$ downward. The MCAT almost always uses $g = 10$ for cleaner arithmetic. Pick a sign convention and be consistent.

• Dropped from rest: $v_i = 0$, $a = -g$.
• Thrown upward: at the peak $v = 0$ instantaneously, but $a = -g$ throughout — a key conceptual point.

Worked Example: Kinematics of a Falling Object

Must know

A package drops from rest off a cliff and hits the ground 3.0 s later. How far does it fall? ($g = 10 \text{ m/s}^2$.)

Using eq. (2) with $v_i = 0$:

$\Delta x = \frac{1}{2}(-10)(3.0)^2 = -45 \text{ m}$

The cliff is 45 m tall. Check with eq. (1): $v_f = (-10)(3.0) = -30 \text{ m/s}$; eq. (3): $v_f^2 = 2(-10)(-45) = 900$, $v_f = -30 \text{ m/s}$. ✓

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Quick check: During free fall from rest, speed increases by $10 \text{ m/s}$ each second. After 4 s, what is the speed?

Answer: $v_f = 0 + (10)(4) = 40 \text{ m/s}$.

Terminal Velocity (one line)

Passage-level

With air resistance, drag grows with speed until it balances gravity; then $\sum F = 0$ and $a = 0$, and the object falls at constant terminal velocity (not maximum acceleration). The MCAT treats projectiles as ideal otherwise.

Two-Dimensional Projectile Motion

Must know

The horizontal and vertical motions are independent. Gravity acts only vertically, so $a_x = 0$ (constant horizontal velocity) while $a_y = -g$. Solve each direction with the 1D equations, linked only by shared time $t$. The path is a symmetric parabola: launch and landing speeds are equal, and rise time equals fall time.

Worked example: Launch at $v_0 = 20 \text{ m/s}$, $\theta = 30°$ from ground level ($g = 10$). Decompose: $v_{0x} \approx 17.3 \text{ m/s}$, $v_{0y} = 10 \text{ m/s}$.

• Time of flight ($\Delta y = 0$): $t = \dfrac{2v_{0y}}{g} = 2.0 \text{ s}$
• Range: $\Delta x = v_{0x}\,t \approx 35 \text{ m}$
• Max height ($v_y = 0$): $h = \dfrac{v_{0y}^2}{2g} = 5.0 \text{ m}$

Sanity check: at the peak only the vertical velocity is zero; horizontal velocity stays $17.3 \text{ m/s}$. ✓

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Common Confusions & Tricks

1. Displacement ≠ Distance (velocity ≠ speed). Round trips, curved paths, and direction changes make these differ. "Average velocity = 0" means the object returned to its start — though it may have traveled far.

2. "Deceleration" ≠ negative acceleration. If an object moves in the −direction and slows, its acceleration is positive (opposing motion). Deceleration just means $\vec{a}$ opposes $\vec{v}$; determine the sign from the direction of $\vec{a}$.

3. At the peak of projectile motion, $v_y = 0$ but $a \neq 0$. Gravity is $-10 \text{ m/s}^2$ at every point, including the top. Only the vertical velocity is zero there.

4. When subtracting vectors, flip the arrow. $\vec{A} - \vec{B}$ adds $-\vec{B}$ (opposite to $\vec{B}$). Don't subtract magnitudes unless both vectors point the same way.

5. Slope is the derivative — use it. Instantaneous velocity = slope of $x$ vs. $t$; instantaneous acceleration = slope of $v$ vs. $t$; displacement = area under $v$ vs. $t$.

6. $g = 10 \text{ m/s}^2$ on the MCAT. The real value is $9.8$, but use $10$ unless a passage specifies otherwise.

7. Round trips at different speeds — don't average the speeds. For equal-distance legs, average speed is the harmonic mean:
$\bar{v} = \frac{2v_1 v_2}{v_1 + v_2}$
The arithmetic average overestimates because you spend more time at the slower speed.

8. Choosing a kinematic equation. Identify which variable is absent from the problem; pick the equation that does not contain it. Equation (3) has no $t$; equation (4) has no $a$.

9. Keep signs consistent. Pick a positive direction at the start and stick with it.

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Key Equations

Equation	Variables & Notes
$A_x = A\cos\theta, \quad A_y = A\sin\theta$	Decompose vector $A$ at angle $\theta$ from +x-axis
$A = \sqrt{A_x^2 + A_y^2}$	Magnitude from components
$\bar{v}_{\text{speed}} = \dfrac{d_{\text{total}}}{\Delta t}$	Average speed; scalar; total path length
$\bar{\vec{v}} = \dfrac{\Delta \vec{r}}{\Delta t}$	Average velocity; vector; net displacement
$\bar{\vec{a}} = \dfrac{\vec{v}_f - \vec{v}_i}{\Delta t}$	Average acceleration; units $\text{m/s}^2$
$v_f = v_i + at$	Eq. 1; use when $\Delta x$ is absent
$\Delta x = v_i t + \tfrac{1}{2}at^2$	Eq. 2; use when $v_f$ is absent
$v_f^2 = v_i^2 + 2a\,\Delta x$	Eq. 3; use when $t$ is absent (most common!)
$\Delta x = \dfrac{v_i + v_f}{2}\cdot t$	Eq. 4; use when $a$ is absent
$g \approx 10 \text{ m/s}^2 \text{ (downward)}$	Free-fall acceleration; use $9.8$ only if specified