MCAT Chemical and Physical Foundations of Biological Systems Flashcards: 4a Kinematics Motion Variables

What this deck covers

This deck focuses on 4a Kinematics Motion Variables, giving you a quick way to review the definitions, rules, and examples that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use these flashcards

Work through these flashcards in short sessions. Try to answer each prompt before flipping the card, then revisit any cards you miss until the explanation feels automatic.

Flashcard 1: Find $v_{0x}$ and $v_{0y}$ if speed is $10\,\text{m/s}$ at $30^\circ$ above horizontal.

Answer: $v_{0x}=10\cos30^\circ$, $v_{0y}=10\sin30^\circ$. Initial velocity components resolve using trigonometry, with cosine for horizontal and sine for vertical relative to the angle.

Flashcard 2: What is the magnitude and direction of gravitational acceleration near Earth’s surface?

Answer: $g \approx 9.8\,\text{m}\,\text{s}^{-2}$ downward. Near Earth's surface, gravity causes constant downward acceleration, approximated as 9.8 m/s² for kinematic calculations.

Flashcard 3: State the constant-acceleration equation relating $v$, $v_0$, $a$, and $t$.

Answer: $v = v_0 + at$. This equation derives from constant acceleration, integrating acceleration to find velocity as initial plus acceleration times time.

Flashcard 4: What is the SI unit of acceleration expressed in base units?

Answer: Meters per second squared: $\text{m}\,\text{s}^{-2}$. Acceleration, as change in velocity over time, has SI units derived from meters per second per second.

Flashcard 5: State the kinematic equations for projectile components (no air resistance) in 2D.

Answer: $x=x_0+v_{0x}t$ and $y=y_0+v_{0y}t-\frac{1}{2}gt^2$. Projectile equations separate into horizontal constant velocity and vertical constant acceleration due to gravity.

Flashcard 6: In projectile motion without air resistance, which velocity component changes: $v_x$ or $v_y$?

Answer: $v_y$ changes; $v_x$ is constant. In projectile motion, gravity affects only the vertical component, leaving horizontal velocity unchanged without air resistance.

Flashcard 7: Find time to reach max height if a ball is thrown upward with $v_0 = 20\,\text{m/s}$ and $a = -g$.

Answer: $t = \frac{v_0}{g} \approx 2.0\,\text{s}$. Time to max height occurs when $v=0$, so $t = \frac{-v_0}{a}$ with $a=-g$, simplifying to $v_0 / g$.

Flashcard 8: Find $a$ if $v_0 = 5\,\text{m/s}$, $v = 1\,\text{m/s}$, and $t = 2\,\text{s}$.

Answer: $a = -2\,\text{m/s}^2$. Acceleration is found from $a = \frac{v - v_0}{t}$, yielding negative value for deceleration in this case.

Flashcard 9: Find $v$ after $t = 3\,\text{s}$ if $v_0 = 2\,\text{m/s}$ and $a = 4\,\text{m/s}^2$.

Answer: $v = 14\,\text{m/s}$. Using $v = v_0 + at$, final velocity increases linearly from initial under constant positive acceleration.

Flashcard 10: What is the definition of distance traveled in one dimension?

Answer: Distance is total path length traveled (nonnegative scalar). Distance measures the total length of the path taken, always positive as a scalar quantity regardless of direction.

Flashcard 11: What is the definition of instantaneous velocity using calculus notation?

Answer: Instantaneous velocity is $v = \frac{dx}{dt}$. Instantaneous velocity is the derivative of position with respect to time, indicating velocity at a specific moment.

Flashcard 12: What is the definition of instantaneous acceleration using calculus notation?

Answer: Instantaneous acceleration is $a = \frac{dv}{dt} = \frac{d^2x}{dt^2}$. Instantaneous acceleration is the derivative of velocity with respect to time, or second derivative of position, showing rate of velocity change.

Flashcard 13: Which graph’s slope equals instantaneous velocity: $x$ vs $t$ or $v$ vs $t$?

Answer: Slope of an $x$ vs $t$ graph equals instantaneous velocity. The slope on a position-time graph represents the rate of change of position, which defines velocity at that instant.

Flashcard 14: Which graph’s slope equals instantaneous acceleration: $v$ vs $t$ or $x$ vs $t$?

Answer: Slope of a $v$ vs $t$ graph equals instantaneous acceleration. The slope on a velocity-time graph indicates the rate of change of velocity, corresponding to acceleration at that point.

Flashcard 15: State the constant-acceleration equation relating $x$, $x_0$, $v_0$, $a$, and $t$.

Answer: $x = x_0 + v_0 t + \frac{1}{2}at^2$. This position equation results from integrating velocity under constant acceleration, combining initial terms and quadratic acceleration effect.

Flashcard 16: State the constant-acceleration equation relating $v^2$, $v_0^2$, $a$, and displacement.

Answer: $v^2 = v_0^2 + 2a(x - x_0)$. This equation eliminates time by combining velocity and position equations, relating speeds squared to acceleration and displacement.

Flashcard 17: State the constant-acceleration equation for displacement using average of $v_0$ and $v$.

Answer: $x - x_0 = \frac{(v_0 + v)}{2}t$. Displacement equals average velocity times time, where average is the mean of initial and final velocities under constant acceleration.

Flashcard 18: Identify the sign of acceleration when an object moves in $+x$ but slows down.

Answer: Acceleration is negative: $a < 0$. When slowing in the positive direction, acceleration opposes motion, resulting in a negative value in the coordinate system.