Representing motion

Multiple Ways to Represent Motion

Motion can be described using five distinct representations, each revealing different information about the same physical situation.

Strong problem-solving starts with a picture or motion diagram. Then translate into graphs or equations as needed.

In a motion diagram, each dot marks the object's position at equal time intervals. Dots spaced far apart mean the object moves fast. Dots close together mean it moves slowly. If spacing increases, the object accelerates. If spacing decreases, the object decelerates.

MisconceptionStudents often confuse "acceleration" with "moving fast." An object can move slowly yet still accelerate if its velocity is changing.

Exam TipAcceleration describes how velocity changes, not how large velocity is.

Kinematic Equations for Constant Acceleration

Three kinematic equations describe one-dimensional motion when acceleration is constant. Each equation connects a different trio of variables, so choosing the right one depends on what is known and unknown.

Key Equations

Velocity as a function of time:

$${v}_{x}={v}_{x0}+{a}_{x}t$$

Position as a function of time:

$$x={x}_{0}+{v}_{x0}t+\frac{1}{2}{a}_{x}{t}^{2}$$

Velocity–position relation (time-independent):

$${v}_{x}^{2}={v}_{x0}^{2}+2{a}_{x}(x-{x}_{0})$$

Variables:

• ${v}_{x}$ = instantaneous velocity at time $t$ (m/s)
• ${v}_{x0}$ = initial velocity (m/s)
• ${a}_{x}$ = constant acceleration (m/s²)
• $t$ = elapsed time (s)
• $x$ = position at time $t$ (m)
• ${x}_{0}$ = initial position (m)

SI units: position in m, velocity in m/s, acceleration in m/s².

ProportionalityIn the position equation, displacement from rest grows as the square of time. So doubling the time quadruples the displacement when ${v}_{x0}=0$.

Reference sheet status: On the reference sheet.

Before using any equation, define the positive direction. A negative result then means the quantity points opposite to that direction. For example, choosing rightward as positive means a leftward velocity appears as a negative ${v}_{x}$.

Each equation omits exactly one variable:

Pick the equation that contains your unknown and whose other variables are all known.

Examiner InsightFRQs often give three known quantities and ask for two unknowns. Identify which equation has only one unknown first, solve it, then use the result in a second equation.

Exam TipWrite "known / unknown" lists before choosing an equation.

Worked Example: Braking Car

Scenario

A car travels at 25 m/s when the driver brakes, producing a constant acceleration of −5.0 m/s². Find the distance the car travels before stopping.

The car's final velocity is zero. Time is not given, so use the time-independent equation.

Equation used

$${v}_{x}^{2}={v}_{x0}^{2}+2{a}_{x}(x-{x}_{0})$$

Rearrange for displacement:

$$x-{x}_{0}=\frac{{v}_{x}^{2}-{v}_{x0}^{2}}{2{a}_{x}}$$

Given

$${v}_{x}=0\text{ m/s}$$

$${v}_{x0}=25\text{ m/s}$$

$${a}_{x}=-5.0{\text{ m/s}}^{2}$$

Working

$$x-{x}_{0}=\frac{(0{)}^{2}-(25{)}^{2}}{2(-5.0)}$$

$$x-{x}_{0}=\frac{-625}{-10}$$

$$x-{x}_{0}=62.5\text{ m}$$

Answer

$$\Delta x=62.5\text{ m}$$

Interpretation

The car needs 62.5 m to stop. The positive result confirms the displacement is in the original direction of motion, which makes physical sense because the car has not yet reversed.

Free-Fall Near Earth's Surface

Near Earth's surface, every object in free fall [motion under gravity alone, with no air resistance] experiences a downward acceleration of magnitude $g\approx 10{\text{ m/s}}^{2}$.

Key Equations

Free-fall acceleration:

$${a}_{y}=\pm g\approx \pm 10{\text{ m/s}}^{2}$$

Variables:

• ${a}_{y}$ = vertical component of acceleration (m/s²)
• $g$ = magnitude of gravitational acceleration near Earth's surface (m/s²)

SI unit: m/s².

Reference sheet status: On the reference sheet (listed as $g\approx 9.8{\text{ m/s}}^{2}$; use $g\approx 10{\text{ m/s}}^{2}$ unless told otherwise).

The sign of ${a}_{y}$ depends on the chosen positive direction. If upward is positive, then ${a}_{y}=-g$. If downward is positive, then ${a}_{y}=+g$. Pick one convention and keep it for the entire problem.

Free-fall applies to objects moving upward, downward, or at their peak. At the peak, velocity is zero but acceleration is still $g$ downward. The velocity is changing at every instant.

The same three kinematic equations apply. Replace ${a}_{x}$ with ${a}_{y}$ and $x$ with $y$.

MisconceptionStudents often believe acceleration is zero at the highest point because velocity is momentarily zero. Acceleration remains $g$ downward throughout the flight.

Exam TipZero velocity does not imply zero acceleration.

Worked Example: Ball Thrown Upward

Scenario

A ball is thrown straight upward at 20 m/s from ground level. Find the maximum height reached.

At maximum height, ${v}_{y}=0$. Choose upward as positive, so ${a}_{y}=-10{\text{ m/s}}^{2}$.

Equation used

$${v}_{y}^{2}={v}_{y0}^{2}+2{a}_{y}(y-{y}_{0})$$

Rearrange:

$$y-{y}_{0}=\frac{{v}_{y}^{2}-{v}_{y0}^{2}}{2{a}_{y}}$$

Given

$${v}_{y}=0\text{ m/s}$$

$${v}_{y0}=20\text{ m/s}$$

$${a}_{y}=-10{\text{ m/s}}^{2}$$

Working

$$y-{y}_{0}=\frac{(0{)}^{2}-(20{)}^{2}}{2(-10)}$$

$$y-{y}_{0}=\frac{-400}{-20}$$

$$y-{y}_{0}=20\text{ m}$$

Answer

$$\Delta y=20.0\text{ m}$$

Interpretation

The ball rises 20 m before momentarily stopping. Gravity then pulls it back down with the same magnitude of acceleration.

Graphs Linking Position, Velocity, and Acceleration

Position–time, velocity–time, and acceleration–time graphs are connected through slopes and areas. Reading these graphs correctly is one of the most tested skills on the AP exam.

Reading graph conventions: the horizontal axis is time ($t$). The vertical axis is the plotted quantity. A positive value means the quantity points in the chosen positive direction. A negative value means it points opposite. A value of zero means the quantity is zero at that instant, not that the object has stopped in every sense.

The derivative $\frac{dx}{dt}$ notation means "the instantaneous rate at which $x$ changes with respect to $t$." On a graph, this equals the slope of the tangent line at that instant. The integral $\int {v}_{x} dt$ means "the accumulated sum of ${v}_{x}$ over a time interval." On a graph, this equals the area between the curve and the time axis.

For constant acceleration, the $x$-vs-$t$ graph is a parabola. The ${v}_{x}$-vs-$t$ graph is a straight line whose slope equals ${a}_{x}$. The ${a}_{x}$-vs-$t$ graph is a horizontal line.

For non-constant acceleration, the ${v}_{x}$-vs-$t$ curve is not straight. The instantaneous acceleration at any point is the slope of the tangent line there. The displacement over an interval is the area under the ${v}_{x}$ curve, which requires integration.

Examiner InsightA common FRQ asks: "Using the graph, determine the displacement between ${t}_{1}$ and ${t}_{2}$." Calculate the area under the $v$-vs-$t$ curve over that interval. If the curve is not linear, set up the integral.

Exam TipAreas below the time axis count as negative displacement.

Calculus Connections: Derivatives and Integrals of Motion

The relationships among position, velocity, and acceleration have precise calculus forms that extend kinematics beyond constant acceleration.

Key Equations

Instantaneous velocity from position:

$${v}_{x}=\frac{dx}{dt}$$

Instantaneous acceleration from velocity:

$${a}_{x}=\frac{d{v}_{x}}{dt}$$

Displacement from velocity:

$$\Delta x=\int_{{t}_{1}}^{{t}_{2}}{v}_{x}(t) dt$$

Change in velocity from acceleration:

$$\Delta {v}_{x}=\int_{{t}_{1}}^{{t}_{2}}{a}_{x}(t) dt$$

Variables:

• ${v}_{x}$ = instantaneous velocity (m/s)
• ${a}_{x}$ = instantaneous acceleration (m/s²)
• $x$ = position (m)
• $t$ = time (s)
• ${t}_{1}$, ${t}_{2}$ = start and end times of the interval (s)

SI units: same as above.

Reference sheet status: On the reference sheet.

Velocity is the time derivative of position. This means velocity at any instant equals the slope of the $x$-vs-$t$ graph at that instant. Acceleration is the time derivative of velocity, so it equals the slope of the ${v}_{x}$-vs-$t$ graph.

Going in reverse: integrating velocity over a time interval gives displacement. Integrating acceleration over a time interval gives the change in velocity. These integrals correspond to the area under the respective curve.

When acceleration is not constant, the constant-acceleration kinematic equations do not apply. Instead, use differentiation and integration directly.

Derivation: Displacement from a Time-Varying Velocity

Starting point: Definition of instantaneous velocity.

$${v}_{x}=\frac{dx}{dt}$$

Step 1 --- Separate variables

Multiply both sides by $dt$ to isolate the position differential.

$$dx={v}_{x}(t) dt$$

Step 2 --- Integrate both sides over the time interval ${t}_{1}$ to ${t}_{2}$

The left side integrates from $x({t}_{1})$ to $x({t}_{2})$. The right side accumulates velocity over time.

$$\int_{x({t}_{1})}^{x({t}_{2})}dx=\int_{{t}_{1}}^{{t}_{2}}{v}_{x}(t) dt$$

Step 3 --- Evaluate the left side

$$x({t}_{2})-x({t}_{1})=\int_{{t}_{1}}^{{t}_{2}}{v}_{x}(t) dt$$

Result

$$\Delta x=\int_{{t}_{1}}^{{t}_{2}}{v}_{x}(t) dt$$

Displacement equals the definite integral of velocity over the chosen time interval, which is the area under the ${v}_{x}$-vs-$t$ curve.

Worked Example: Non-Constant Velocity

Scenario

An object's velocity is given by ${v}_{x}(t)=4{t}^{2}-2t$ (in m/s). Find the displacement from $t=1.0$ s to $t=3.0$ s.

Step-by-step:

Set up the integral:

$$\Delta x=\int_{1.0}^{3.0}(4{t}^{2}-2t) dt$$

Evaluate using the power rule:

$$\Delta x={\left[\frac{4{t}^{3}}{3}-{t}^{2}\right]}_{1.0}^{3.0}$$

Substitute the upper limit:

$$\frac{4(3.0{)}^{3}}{3}-(3.0{)}^{2}=\frac{108}{3}-9.0=36.0-9.0=27.0$$

Substitute the lower limit:

$$\frac{4(1.0{)}^{3}}{3}-(1.0{)}^{2}=\frac{4}{3}-1.0=0.333$$

Subtract:

$$\Delta x=27.0-0.333=26.7\text{ m}$$

Answer

$$\Delta x\approx 26.7\text{ m}$$

Interpretation

The object moves 26.7 m in the positive direction between $t=1.0$ s and $t=3.0$ s. The positive result confirms net displacement is in the positive direction.