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.

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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: 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 2: 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 3: 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 4: 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 5: 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 6: 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 7: 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 8: 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.

Flashcard 9: 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 10: 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 11: 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: 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.