Kinematics and Motion Variables (4A) - MCAT Chemical and Physical Foundations of Biological Systems
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Find $v_{0x}$ and $v_{0y}$ if speed is $10,\text{m/s}$ at $30^\circ$ above horizontal.
Find $v_{0x}$ and $v_{0y}$ if speed is $10,\text{m/s}$ at $30^\circ$ above horizontal.
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$v_{0x}=10\cos30^\circ$, $v_{0y}=10\sin30^\circ$. Initial velocity components resolve using trigonometry, with cosine for horizontal and sine for vertical relative to the angle.
$v_{0x}=10\cos30^\circ$, $v_{0y}=10\sin30^\circ$. Initial velocity components resolve using trigonometry, with cosine for horizontal and sine for vertical relative to the angle.
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What is the magnitude and direction of gravitational acceleration near Earth’s surface?
What is the magnitude and direction of gravitational acceleration near Earth’s surface?
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$g \approx 9.8,\text{m},\text{s}^{-2}$ downward. Near Earth's surface, gravity causes constant downward acceleration, approximated as 9.8 m/s² for kinematic calculations.
$g \approx 9.8,\text{m},\text{s}^{-2}$ downward. Near Earth's surface, gravity causes constant downward acceleration, approximated as 9.8 m/s² for kinematic calculations.
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State the constant-acceleration equation relating $v$, $v_0$, $a$, and $t$.
State the constant-acceleration equation relating $v$, $v_0$, $a$, and $t$.
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$v = v_0 + at$. This equation derives from constant acceleration, integrating acceleration to find velocity as initial plus acceleration times time.
$v = v_0 + at$. This equation derives from constant acceleration, integrating acceleration to find velocity as initial plus acceleration times time.
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What is the SI unit of acceleration expressed in base units?
What is the SI unit of acceleration expressed in base units?
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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.
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.
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State the kinematic equations for projectile components (no air resistance) in 2D.
State the kinematic equations for projectile components (no air resistance) in 2D.
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$x=x_0+v_{0x}t$ 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.
$x=x_0+v_{0x}t$ 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.
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In projectile motion without air resistance, which velocity component changes: $v_x$ or $v_y$?
In projectile motion without air resistance, which velocity component changes: $v_x$ or $v_y$?
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$v_y$ changes; $v_x$ is constant. In projectile motion, gravity affects only the vertical component, leaving horizontal velocity unchanged without air resistance.
$v_y$ changes; $v_x$ is constant. In projectile motion, gravity affects only the vertical component, leaving horizontal velocity unchanged without air resistance.
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Find time to reach max height if a ball is thrown upward with $v_0 = 20,\text{m/s}$ and $a = -g$.
Find time to reach max height if a ball is thrown upward with $v_0 = 20,\text{m/s}$ and $a = -g$.
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$t = \frac{v_0}{g} \approx 2.0,\text{s}$. Time to max height occurs when $v=0$, so $t = \frac{-v_0}{a}$ with $a=-g$, simplifying to $v_0 / g$.
$t = \frac{v_0}{g} \approx 2.0,\text{s}$. Time to max height occurs when $v=0$, so $t = \frac{-v_0}{a}$ with $a=-g$, simplifying to $v_0 / g$.
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Find $a$ if $v_0 = 5,\text{m/s}$, $v = 1,\text{m/s}$, and $t = 2,\text{s}$.
Find $a$ if $v_0 = 5,\text{m/s}$, $v = 1,\text{m/s}$, and $t = 2,\text{s}$.
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$a = -2,\text{m/s}^2$. Acceleration is found from $a = \frac{v - v_0}{t}$, yielding negative value for deceleration in this case.
$a = -2,\text{m/s}^2$. Acceleration is found from $a = \frac{v - v_0}{t}$, yielding negative value for deceleration in this case.
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Find $v$ after $t = 3,\text{s}$ if $v_0 = 2,\text{m/s}$ and $a = 4,\text{m/s}^2$.
Find $v$ after $t = 3,\text{s}$ if $v_0 = 2,\text{m/s}$ and $a = 4,\text{m/s}^2$.
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$v = 14,\text{m/s}$. Using $v = v_0 + at$, final velocity increases linearly from initial under constant positive acceleration.
$v = 14,\text{m/s}$. Using $v = v_0 + at$, final velocity increases linearly from initial under constant positive acceleration.
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What is the definition of distance traveled in one dimension?
What is the definition of distance traveled in one dimension?
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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.
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.
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What is the definition of instantaneous velocity using calculus notation?
What is the definition of instantaneous velocity using calculus notation?
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Instantaneous velocity is $v = \frac{dx}{dt}$. Instantaneous velocity is the derivative of position with respect to time, indicating velocity at a specific moment.
Instantaneous velocity is $v = \frac{dx}{dt}$. Instantaneous velocity is the derivative of position with respect to time, indicating velocity at a specific moment.
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What is the definition of instantaneous acceleration using calculus notation?
What is the definition of instantaneous acceleration using calculus notation?
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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.
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.
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Which graph’s slope equals instantaneous velocity: $x$ vs $t$ or $v$ vs $t$?
Which graph’s slope equals instantaneous velocity: $x$ vs $t$ or $v$ vs $t$?
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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.
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.
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Which graph’s slope equals instantaneous acceleration: $v$ vs $t$ or $x$ vs $t$?
Which graph’s slope equals instantaneous acceleration: $v$ vs $t$ or $x$ vs $t$?
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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.
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.
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State the constant-acceleration equation relating $x$, $x_0$, $v_0$, $a$, and $t$.
State the constant-acceleration equation relating $x$, $x_0$, $v_0$, $a$, and $t$.
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$x = x_0 + v_0 t + \frac{1}{2}at^2$. This position equation results from integrating velocity under constant acceleration, combining initial terms and quadratic acceleration effect.
$x = x_0 + v_0 t + \frac{1}{2}at^2$. This position equation results from integrating velocity under constant acceleration, combining initial terms and quadratic acceleration effect.
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State the constant-acceleration equation relating $v^2$, $v_0^2$, $a$, and displacement.
State the constant-acceleration equation relating $v^2$, $v_0^2$, $a$, and displacement.
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$v^2 = v_0^2 + 2a(x - x_0)$. This equation eliminates time by combining velocity and position equations, relating speeds squared to acceleration and displacement.
$v^2 = v_0^2 + 2a(x - x_0)$. This equation eliminates time by combining velocity and position equations, relating speeds squared to acceleration and displacement.
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State the constant-acceleration equation for displacement using average of $v_0$ and $v$.
State the constant-acceleration equation for displacement using average of $v_0$ and $v$.
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$x - x_0 = \frac{(v_0 + v)}{2}t$. Displacement equals average velocity times time, where average is the mean of initial and final velocities under constant acceleration.
$x - x_0 = \frac{(v_0 + v)}{2}t$. Displacement equals average velocity times time, where average is the mean of initial and final velocities under constant acceleration.
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Identify the sign of acceleration when an object moves in $+x$ but slows down.
Identify the sign of acceleration when an object moves in $+x$ but slows down.
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Acceleration is negative: $a < 0$. When slowing in the positive direction, acceleration opposes motion, resulting in a negative value in the coordinate system.
Acceleration is negative: $a < 0$. When slowing in the positive direction, acceleration opposes motion, resulting in a negative value in the coordinate system.
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