2.1 Translational Kinematics, Projectile Motion & Newton's Laws

2.1 Translational Kinematics, Projectile Motion & Newton's Laws

Mechanics forms the physical foundation for understanding human physiology, musculoskeletal biomechanics, and medical diagnostics. On the MCAT, kinematics and dynamics problems rarely appear in pure abstraction; instead, the AAMC embeds translational motion within biological frameworks such as running mechanics, tendon tension, bone fracture limits, and centrifugation of cellular homogenates. Mastering these principles requires both mathematical fluency with vector equations and the conceptual ability to translate anatomical structures into mechanical free-body diagrams.

---

The Big 4 Kinematic Equations

Translational kinematics describes the motion of objects without reference to the forces causing that motion. On the MCAT, kinematic equations are strictly valid only when acceleration is constant (uniform) throughout the interval of interest. If acceleration varies with time or displacement (such as during variable drag), calculus is required, and the standard kinematic equations cannot be applied directly.

There are five primary kinematic variables:

1. $\Delta x$ (or $\Delta y$): Displacement (m)
2. $v_0$: Initial velocity at time $t = 0$ (m/s)
3. $v$: Final velocity at time $t$ (m/s)
4. $a$: Uniform acceleration (m/s²)
5. $t$: Elapsed time (s)

Every kinematic equation links four of these five variables, leaving exactly one variable missing. The fastest method to solve an MCAT kinematics item is to identify the unmentioned variable and select the equation that omits it.

Kinematic Equations and Missing Variable Strategy

Vertical Free-Fall Conventions

Near Earth's surface, all free-falling objects experience a downward gravitational acceleration of $g \approx 9.8\text{ m/s}^2$ (conveniently approximated as $10\text{ m/s}^2$ on the MCAT to streamline mental math). Always define a consistent coordinate system before substituting numbers:

• If upward is chosen as positive (+), then acceleration due to gravity is negative: $a = -g = -10\text{ m/s}^2$. An object launched upward with initial velocity $+v_0$ decelerates until its instantaneous vertical velocity reaches $v_y = 0$ at maximum height.
• If downward is chosen as positive (+) (convenient for objects dropped from rest), $a = +g = +10\text{ m/s}^2$, and displacement $\Delta y$ is positive downward.

---

Projectile Motion in Biomechanics

Projectile motion occurs when an object is launched into two-dimensional space and moves solely under the influence of gravity (neglecting aerodynamic drag). The governing principle of projectile mechanics is the independence of orthogonal motion components: the horizontal and vertical motions do not influence each other.

Vector Decomposition

Given an initial launch speed $v_0$ at an angle $\theta$ relative to the horizontal:

• Horizontal component: $v_{0x} = v_0 \cos\theta$
• Vertical component: $v_{0y} = v_0 \sin\theta$

Horizontal Motion ($x$-axis)

Because gravity acts strictly downward, there is no horizontal acceleration ($a_x = 0$) in the absence of air resistance. Consequently, horizontal velocity remains constant throughout the entire trajectory:

$$v_x(t) = v_{0x} = v_0 \cos\theta$$ $$x(t) = (v_0 \cos\theta) t$$

Vertical Motion ($y$-axis)

Vertical motion is governed by uniform downward acceleration ($a_y = -g$):

$$v_y(t) = v_0 \sin\theta - g t$$ $$y(t) = (v_0 \sin\theta) t - \frac{1}{2} g t^2$$ $$v_y^2 = (v_0 \sin\theta)^2 - 2 g \Delta y$$

Trajectory Symmetry and Derived Metrics

For a projectile launched and landing at the identical vertical elevation ($y = 0$):

1. Time to apex ($t_{\text{apex}}$): At the highest point, vertical velocity momentarily reaches zero ($v_y = 0$): $$0 = v_0 \sin\theta - g t_{\text{apex}} \implies t_{\text{apex}} = \frac{v_0 \sin\theta}{g}$$
2. Total flight time ($t_{\text{total}}$): By parabolic symmetry, the ascent time equals the descent time: $$t_{\text{total}} = 2 t_{\text{apex}} = \frac{2 v_0 \sin\theta}{g}$$
3. Maximum apex height ($H$): $$0 = (v_0 \sin\theta)^2 - 2 g H \implies H = \frac{(v_0 \sin\theta)^2}{2 g}$$
4. Horizontal range ($R$): $$R = v_x \cdot t_{\text{total}} = (v_0 \cos\theta) \left(\frac{2 v_0 \sin\theta}{g}\right) = \frac{v_0^2 \sin(2\theta)}{g}$$ Maximum horizontal range occurs when $\sin(2\theta) = 1$, which corresponds to a launch angle of $\theta = 45^\circ$.

In animal locomotion, such as a jumping frog or an athlete performing a vertical leap, mechanical power generated during leg extension establishes $v_0$. Maximizing height requires directing all takeoff velocity vertically ($\theta = 90^\circ$), whereas maximizing forward jumping distance compromises between horizontal velocity and flight time (near $\theta = 45^\circ$).

---

Newton's Three Laws of Motion

Newtonian dynamics connects kinematics to the physical forces driving or resisting motion.

Newton's First Law (Law of Inertia)

An object remains at rest or continues to move at a constant velocity in an inertial reference frame unless acted upon by a net external force:

$$\sum \vec{F} = 0 \iff \vec{a} = 0$$

Inertia is quantified by mass ($m$). An object in translational equilibrium has zero net force ($\sum F_x = 0$ and $\sum F_y = 0$). Note that zero acceleration does not imply zero velocity; it implies that velocity is constant in both magnitude and direction.

Newton's Second Law (Fundamental Law of Dynamics)

The net force acting on a body is equal to the product of its mass and its acceleration:

$$\sum \vec{F} = m \vec{a}$$

Force and acceleration are collinear vectors. On the MCAT, Newton's second law is evaluated by decomposing forces into perpendicular components:

$$\sum F_x = m a_x, \quad \sum F_y = m a_y$$

Newton's Third Law (Action-Reaction)

When body A exerts a force on body B, body B simultaneously exerts a force on body A that is equal in magnitude and opposite in direction:

$$\vec{F}_{A \text{ on } B} = -\vec{F}_{B \text{ on } A}$$

Critical MCAT Pitfall: Action-reaction pairs never act on the same body. For a person standing motionless on a floor, Earth exerts a downward gravitational force on the person ($F_g$), and the floor exerts an upward normal force on the person ($N$). While $N = F_g$ in static equilibrium, they are not an action-reaction pair because both forces act on the same body (the person). The true Newton's third law reaction to Earth's gravity on the person is the gravitational pull of the person on Earth; the reaction to the floor pushing up on the feet is the feet pushing down on the floor.

---