Kinematics

Kinematics: Describing Motion

Part 1 of 4 — Motion Concepts & Graphs

Kinematics is the language of motion — it describes how things move without asking why. On the MCAT, kinematics questions reward precise vocabulary and fast graph reading far more than heavy algebra.

Scalars vs. Vectors

• Distance is the total path length traveled — it never decreases.
• Displacement $\Delta x = x_f - x_0$ is the straight-line change in position. Walk one full lap around a track: distance $= 400\;\text{m}$, displacement $= 0$.

Velocity and Speed

$\bar{v} = \frac{\Delta x}{\Delta t} \quad \text{(average velocity, uses displacement)}$

$\text{average speed} = \frac{\text{total distance}}{\text{total time}}$

Average speed is always greater than or equal to the magnitude of average velocity; they are equal only when the motion never reverses direction.

Acceleration

$\bar{a} = \frac{\Delta v}{\Delta t}$

Acceleration is any change in velocity — speeding up, slowing down, or turning. A car rounding a curve at constant speed is accelerating because the direction of its velocity changes.

Sign convention: "deceleration" simply means acceleration pointing opposite to the velocity. An object moving in the $+x$ direction and slowing down has a negative acceleration.

Reading Motion Graphs (High Yield)

The MCAT loves graph questions because they test understanding, not computation. Memorize this table:

Key readings:

• A horizontal line on an $x$-$t$ graph means the object is at rest.
• A straight, sloped line on an $x$-$t$ graph means constant velocity.
• A curved $x$-$t$ graph means the velocity is changing (acceleration).
• On a $v$-$t$ graph, crossing the time axis means the object reverses direction.
• Area below the time axis on a $v$-$t$ graph counts as negative displacement.

Worked Example — Average Speed vs. Average Velocity

A nurse walks $300\;\text{m}$ east down a hospital corridor in $40\;\text{s}$, realizes she forgot a chart, and walks $100\;\text{m}$ back west in $10\;\text{s}$.

Step 1 — Total distance and total time.

$d = 300 + 100 = 400\;\text{m}, \qquad t = 40 + 10 = 50\;\text{s}$

Step 2 — Average speed.

$\text{speed} = \frac{400}{50} = 8\;\text{m/s}$

Step 3 — Displacement and average velocity.

$\Delta x = 300 - 100 = 200\;\text{m east}, \qquad \bar{v} = \frac{200}{50} = 4\;\text{m/s east}$

The average speed ($8\;\text{m/s}$) exceeds the magnitude of the average velocity ($4\;\text{m/s}$) because the path doubled back on itself.

Motion Concepts & Graphs 🎯

Key Takeaways — Part 1

• Distance and speed are scalars; displacement, velocity, and acceleration are vectors
• Average speed $\geq$ |average velocity|, with equality only for non-reversing straight-line motion
• Acceleration is any change in velocity — including turning at constant speed
• Graphs: slope of $x$-$t$ is velocity; slope of $v$-$t$ is acceleration; area under $v$-$t$ is displacement
• Zero velocity does NOT imply zero acceleration (top of a throw)

Kinematics: Describing Motion

Part 2 of 4 — The Kinematic Equation Toolkit

When acceleration is constant, five equations connect the variables $v_0$, $v$, $a$, $t$, and $\Delta x$. Each equation is missing exactly one variable — pick the one missing the quantity you neither know nor want.

The Big 5

MCAT strategy: the no-time equation $v^2 = v_0^2 + 2a\Delta x$ is the single most used — MCAT problems love giving distances and hiding time.

Free Fall (use $g = 10\;\text{m/s}^2$)

Free fall is just constant acceleration with $a = g = 10\;\text{m/s}^2$ downward:

• Dropped from rest: $v = gt$ and $\Delta y = \tfrac{1}{2}g t^2 = 5t^2$
• Speed grows by $10\;\text{m/s}$ every second: after $1, 2, 3\;\text{s}$ the speed is $10, 20, 30\;\text{m/s}$
• Distances fallen from rest: $5, 20, 45\;\text{m}$ after $1, 2, 3\;\text{s}$

Symmetry of Vertical Throws

For a ball thrown straight up and caught at the same height:

• Time up $=$ time down: $t_{up} = \frac{v_0}{g}$
• Max height: $H = \frac{v_0^2}{2g}$
• It returns with the same speed it left with, pointing down
• At the top: $v = 0$ but $a = g$ (never zero)

Worked Example — A Vertical Throw

A ball is thrown straight up at $30\;\text{m/s}$. Using $g = 10\;\text{m/s}^2$, find the time to the top, the maximum height, and the total flight time.

Step 1 — Time to the top. At the peak $v = 0$:

$0 = 30 - 10t \Rightarrow t_{up} = 3\;\text{s}$

Step 2 — Maximum height. Use the no-time equation:

$0 = (30)^2 - 2(10)H \Rightarrow H = \frac{900}{20} = 45\;\text{m}$

Step 3 — Total flight time. By symmetry:

$T = 2t_{up} = 6\;\text{s}$

and the ball lands moving $30\;\text{m/s}$ downward.

Worked Example — Constant Acceleration from Rest

A sprinter accelerates from rest at $2\;\text{m/s}^2$ for $6\;\text{s}$.

$v = 0 + (2)(6) = 12\;\text{m/s}$

$\Delta x = 0 + \tfrac{1}{2}(2)(6)^2 = (1)(36) = 36\;\text{m}$

Check with the average-velocity equation: $\Delta x = \tfrac{1}{2}(0 + 12)(6) = 36\;\text{m}$ ✓ — averaging initial and final velocity is often the fastest MCAT route when acceleration is constant.

Kinematic Equations 🎯

Key Takeaways — Part 2

• Pick the Big 5 equation missing the variable you don't know and don't need
• $v^2 = v_0^2 + 2a\Delta x$ is the workhorse when time isn't given
• Free fall with $g = 10$: speeds go $10, 20, 30\;\text{m/s}$ and distances $5, 20, 45\;\text{m}$ after $1, 2, 3\;\text{s}$
• Vertical throws are symmetric: $t_{up} = v_0/g$, $H = v_0^2/2g$, return speed = launch speed
• $\Delta x = \tfrac{1}{2}(v_0 + v)t$ (average velocity) is often the fastest no-calculator path

Kinematics: Describing Motion

Part 3 of 4 — Projectiles & Free Fall in Practice

The One Rule of Projectile Motion

Horizontal and vertical motions are completely independent. Gravity acts only vertically.

Consequences the MCAT tests over and over:

• A bullet fired horizontally and one dropped from the same height hit the ground at the same time
• The horizontal speed never changes the fall time
• At the apex, the velocity is purely horizontal (not zero, unless launched straight up)
• Range is maximized at $45^\circ$; complementary angles (e.g., $30^\circ$ and $60^\circ$) give the same range

MCAT-Friendly Trig Values

The MCAT hands you triangles you can do in your head. Memorize:

$\sin 30^\circ = 0.5, \quad \cos 30^\circ \approx 0.87$

$\sin 37^\circ \approx 0.6, \quad \cos 37^\circ \approx 0.8 \quad \text{(the 3-4-5 triangle)}$

$\sin 45^\circ = \cos 45^\circ \approx 0.7$

$\sin 53^\circ \approx 0.8, \quad \cos 53^\circ \approx 0.6$

Worked Example — Angled Launch (3-4-5 numbers)

A projectile is launched at $50\;\text{m/s}$ at $37^\circ$ above the horizontal ($g = 10\;\text{m/s}^2$).

Step 1 — Resolve the launch velocity.

$v_x = 50\cos 37^\circ = 50(0.8) = 40\;\text{m/s}$

$v_{y0} = 50\sin 37^\circ = 50(0.6) = 30\;\text{m/s}$

Step 2 — Time to apex and max height.

$t_{up} = \frac{30}{10} = 3\;\text{s}, \qquad H = \frac{(30)^2}{2(10)} = \frac{900}{20} = 45\;\text{m}$

Step 3 — Total time and range.

$T = 2(3) = 6\;\text{s}, \qquad R = v_x T = (40)(6) = 240\;\text{m}$

Worked Example — Horizontal Launch

A ball leaves a table $45\;\text{m}$ high (a cliff, really) moving horizontally at $20\;\text{m/s}$.

Fall time (vertical motion starts from $v_{y0} = 0$):

$45 = 5t^2 \Rightarrow t^2 = 9 \Rightarrow t = 3\;\text{s}$

Horizontal distance: $x = (20)(3) = 60\;\text{m}$

Landing velocity: $v_x = 20\;\text{m/s}$ (unchanged), $v_y = (10)(3) = 30\;\text{m/s}$, so the landing speed is

$v = \sqrt{20^2 + 30^2} = \sqrt{400 + 900} = \sqrt{1300} \approx 36\;\text{m/s}$

On the MCAT, estimate: $\sqrt{1300}$ is a bit more than $\sqrt{1225} = 35$.

Projectiles & Free Fall 🎯

Key Takeaways — Part 3

• Split every projectile into independent $x$ (constant velocity) and $y$ (free fall) motions
• Fall time comes from the vertical equation alone; range = $v_x \times$ (total time)
• Memorize the 3-4-5 values: $\sin 37^\circ \approx 0.6$, $\cos 37^\circ \approx 0.8$
• Apex velocity is horizontal, not zero; complementary launch angles share a range
• Sanity-check answers: a fall from bed height lands at about $6\;\text{m/s}$, not $36$

Kinematics: Describing Motion

Part 4 of 4 — MCAT Integration: Experiments, Estimation & Passage Strategy

On the real exam, kinematics rarely arrives as a bare "plug into the Big 5" problem. It shows up inside experimental passages — reaction-time studies, motion-sensor traces, gait analysis — where the physics is easy but buried in data.

How Kinematics Appears in Passages

• Reaction-time experiments: a ruler or light is dropped and the subject responds. Distance fallen from rest obeys $d = 5t^2$ (with $g = 10\;\text{m/s}^2$), so the fall distance is a clock: $t = \sqrt{d/5}$.
• Motion-sensor / video-frame data: you get a table of position or velocity vs. time. Slope and area, not equations, answer these fastest.
• Relative velocity: velocities along one line simply add or subtract. A swimmer at $v_{swim}$ relative to the water in a current $v_{current}$ moves at $v_{swim} \pm v_{current}$ relative to the shore.
• Air resistance (qualitative only): drag grows with speed. A falling object approaches terminal velocity, where drag balances weight, so $a = 0$ while $v$ is at its maximum — the exam loves this "zero acceleration, nonzero velocity" inversion of the apex fact from Part 2.

Estimation Beats Algebra

The MCAT gives no calculator, so it gives clean numbers. Train these reflexes:

• $\sqrt{0.04} = 0.2$, $\sqrt{0.09} = 0.3$ — reaction times are fractions of a second
• Free-fall distances from rest: $5, 20, 45, 80\;\text{m}$ at $1, 2, 3, 4\;\text{s}$
• Round-trip time for a throw at $v_0$: $T = 2v_0/g = v_0/5$

Answer-choice triage: before computing, scan the options. If they differ by powers of ten, dimensional analysis or a one-digit estimate settles it. If two options are $x$ and $x^2$ (like $6$ and $36$), the test is checking whether you took the square root.

Worked Example — A Reaction-Time Passage

Researchers measure driver reaction time with a drop test: a vertical ruler is released without warning, and the subject pinches it as fast as possible. The distance the ruler falls before capture is recorded. Trials for one subject (well rested vs. sleep deprived) average $0.20\;\text{m}$ and $0.45\;\text{m}$ respectively. Use $g = 10\;\text{m/s}^2$.

Step 1 — Convert distance to time (rested). The ruler falls from rest:

$d = \tfrac{1}{2}gt^2 = 5t^2 \Rightarrow 0.20 = 5t^2 \Rightarrow t^2 = 0.04 \Rightarrow t = 0.2\;\text{s}$

Step 2 — Sleep-deprived trial.

$0.45 = 5t^2 \Rightarrow t^2 = 0.09 \Rightarrow t = 0.3\;\text{s}$

Sleep deprivation added $0.1\;\text{s}$ — a $50\%$ increase in reaction time, even though the fall distance more than doubled. Distance is not proportional to time ($d \propto t^2$): a classic passage trap is to ask about the ratio of times, hoping you answer with the ratio of distances.

Step 3 — Translate to the road. At highway speed, $30\;\text{m/s}$, the extra $0.1\;\text{s}$ of reaction time adds

$\Delta x = (30)(0.1) = 3\;\text{m}$

of travel before braking even begins — the passage will typically close the loop with exactly this kind of constant-velocity distance question.

Passage strategy recap: identify which phase of motion each question targets (constant velocity before the response vs. free fall of the ruler), pull only the numbers that phase needs, and keep the phases separate — the same discipline as splitting projectile motion into $x$ and $y$.

Kinematics Integration & Passage Strategy 🎯

Key Takeaways — Part 4

• Reaction-time drop tests are free fall in disguise: $t = \sqrt{d/5}$ with $g = 10\;\text{m/s}^2$
• On data tables and graphs, reach for slope and area before reaching for equations
• Relative velocity along a line is simple addition and subtraction of signed velocities
• Terminal velocity: maximum speed with zero acceleration — drag balances weight
• Triage answer choices first: powers of ten, missing square roots, and $x$ vs. $x^2$ pairs give answers away
• Keep motion phases separate (constant-velocity travel vs. free fall), just like $x$ and $y$ in projectiles