Unit 1: Kinematics

Describing Motion: Position, Time, and Reference Frames

Kinematics is the part of mechanics that describes how objects move without considering the forces that cause the motion (forces come later). In AP Physics C: Mechanics, kinematics is calculus-based, so you are expected to connect motion to derivatives and integrals, not just memorize constant-acceleration equations.

A useful way to think about kinematics is that you are building a model: choose a coordinate system, define variables like position and time, and then use mathematical relationships to describe motion. If the setup is clear, the math becomes much easier and sign errors become less common.

Position, coordinate systems, and reference frames

Position tells you where an object is located relative to an origin. In one dimension, position is often written as a function of time:

x(t)

In two dimensions, position is a vector:

\vec r(t)

A reference frame is the viewpoint from which positions and times are measured. In this unit (and generally in AP Physics C unless stated otherwise), reference frames are treated as inertial (not accelerating). This matters later for Newton’s laws, but even in kinematics it affects what it means to be “at rest.”

When choosing a coordinate system, decide:

• Where the origin is
• Which direction is positive
• In 2D, how the axes are oriented

The choice is arbitrary, but it must stay consistent throughout the solution.

Distance vs displacement

Distance is the total length of the path traveled and is always nonnegative.

Displacement is the signed change in position from initial to final:

\Delta x = x_f - x_i

An object can travel a nonzero distance but have zero displacement (for example, moving right and then returning to the start). Velocity and acceleration connect to displacement (signed change in position), not distance.

Time intervals

Time is a scalar variable:

t

A time interval is:

\Delta t = t_f - t_i

In most kinematics problems, \Delta t is positive. A negative value usually means the initial and final labels were swapped.

Speed vs velocity (basic definitions)

Speed is the rate of covering distance.

Velocity is the rate of changing displacement, so it includes direction (it is a vector quantity in general).

Average speed is:

\text{average speed} = \frac{\text{total distance}}{\Delta t}

Average velocity is:

v_{avg} = \frac{\Delta x}{\Delta t}

Average speed and average velocity are often different because distance and displacement are different.

Notation you will see (and how to interpret it)

A subtle but important point: the symbol v can mean speed (a magnitude) in some contexts, but in AP Physics C it is often used as velocity in 1D, where the sign carries direction.

Example: choosing a coordinate system wisely

Suppose an elevator moves upward from the lobby to the 10th floor.

• Convenient choice: take up as positive, origin at the lobby. Then the elevator’s motion has positive displacement and (during upward motion) positive velocity.
• If you choose down as positive, the physics is still correct if you stay consistent, but many students accidentally mix conventions (for example, treating upward motion as positive while also treating gravity as positive).

The physics does not change, only the signs do. A good coordinate choice reduces the number of sign flips you must track.

Exam Focus

• Typical question patterns:
  • Define a coordinate system and interpret the meaning of negative position, velocity, or acceleration.
  • Distinguish displacement from distance using a motion description or a graph.
  • Identify initial and final values correctly from wording (“from A to B,” “returns to start,” etc.).
  • Compare average speed vs average velocity for a described trip.
• Common mistakes:
  • Treating distance traveled as \Delta x.
  • Changing the positive direction mid-solution.
  • Confusing “position is negative” with “moving in the negative direction.”

Velocity: Average, Instantaneous, and the Derivative

Velocity captures how fast and in what direction position changes. In AP Physics C, the key idea is that velocity is a derivative, which lets you solve many problems even when acceleration is not constant.

Average velocity

Average velocity over a time interval is displacement divided by elapsed time:

v_{avg} = \frac{\Delta x}{\Delta t}

Average velocity depends only on starting and ending positions, not on the path details.

Instantaneous velocity and the derivative

Instantaneous velocity is the limit of average velocity as the time interval shrinks:

v = \lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t}

If position is differentiable, then:

v(t) = \frac{dx}{dt}

This means velocity is the slope of the position-versus-time graph at each instant.

Speed vs velocity

Speed ignores direction. In 1D, speed is the magnitude of velocity:

\text{speed} = |v|

An object can have v = 0 at an instant (for example, at the top of a trajectory) while still having nonzero acceleration.

Interpreting velocity on graphs

From a position graph x(t):

• Positive slope means v > 0.
• Negative slope means v < 0.
• Steeper slope means larger velocity magnitude.
• A horizontal tangent means v = 0 at that instant.

From a velocity graph v(t):

• The sign of v tells direction.
• The value of v is the instantaneous velocity.

Example 1: average vs instantaneous velocity (calculus)

An object moves along a line with position:

x(t) = 3t^2 - 2t + 5

1) Instantaneous velocity:

v(t) = \frac{dx}{dt} = 6t - 2

2) Average velocity from t = 1 to t = 3:

x(1) = 6

x(3) = 26

v_{avg} = \frac{26 - 6}{3 - 1} = 10

Average velocity is one number for an interval, while instantaneous velocity is a function of time.

Example 2: speed from a position function

An object’s position obeys:

x(t) = \sin(t)

How fast is it moving at:

t = \frac{\pi}{2}

Velocity is the derivative:

v(t) = \cos(t)

Evaluate:

v\left(\frac{\pi}{2}\right) = 0

So at that instant the object’s velocity is zero.

Example 3: turning points and velocity

If a position function has a local maximum or minimum, its slope there is zero, so:

v = 0

This does not mean the object stays stopped; it means it is momentarily not changing position and may reverse direction immediately.

Exam Focus

• Typical question patterns:
  • Differentiate a position function to find v(t) and evaluate at a specified time.
  • Use a tangent slope on an x(t) graph to estimate instantaneous velocity.
  • Interpret intervals of positive or negative velocity from a graph.
  • Distinguish average speed from average velocity in word problems.
• Common mistakes:
  • Using average velocity when the question asks for instantaneous velocity (or vice versa).
  • Confusing speed with velocity, especially when velocity is negative.
  • Forgetting units: if x is meters and t is seconds, then v is meters per second.

Acceleration: The Derivative of Velocity and the Shape of Motion

Acceleration describes how velocity changes. It determines whether you speed up, slow down, or change direction. Acceleration is a vector quantity (it has magnitude and direction). If an object moves with constant velocity, then its acceleration is zero.

Average acceleration

Average acceleration over an interval is:

a_{avg} = \frac{\Delta v}{\Delta t}

where:

\Delta v = v_f - v_i

Instantaneous acceleration and the derivative

Instantaneous acceleration is:

a = \lim_{\Delta t \to 0} \frac{\Delta v}{\Delta t}

In calculus form:

a(t) = \frac{dv}{dt}

Since:

v(t) = \frac{dx}{dt}

acceleration is also:

a(t) = \frac{d^2x}{dt^2}

So the curvature (concavity) of an x(t) graph relates directly to acceleration.

Acceleration sample problem (average acceleration)

A car travels along a straight highway at 20 m/s. Three seconds later, its speed is 32 m/s. Find its average acceleration.

Change in velocity is:

\Delta v = 32 - 20 = 12

Average acceleration:

a_{avg} = \frac{12}{3} = 4

Sign of acceleration vs “speeding up” in 1D

“Positive acceleration” does not always mean “speeding up.” In 1D:

• If v > 0 and a > 0, speed increases.
• If v < 0 and a < 0, speed increases.
• If v > 0 and a < 0, speed decreases.
• If v < 0 and a > 0, speed decreases.

A reliable rule is: speed increases when velocity and acceleration have the same sign (in 1D).

Constant (uniform) acceleration and why it is special

Uniformly accelerated motion means acceleration remains constant. Key points:

• The acceleration remains constant throughout the motion.
• The velocity changes at a constant rate.
• The displacement is directly proportional to the square of time elapsed.

When acceleration is constant, integrating the derivative relationships gives the standard constant-acceleration equations:

v = v_0 + at

x = x_0 + v_0 t + \frac{1}{2}at^2

An additional useful form eliminates time:

v^2 = v_0^2 + 2a(x - x_0)

Another commonly used constant-acceleration relationship is:

x - x_0 = vt - \frac{1}{2}at^2

Equations of motion with alternate symbols

You may also see the same constant-acceleration equations written with different symbols:

• initial velocity written as u instead of v_0
• displacement written as s instead of \Delta x

For example:

v = u + at

s = ut + \frac{1}{2}at^2

v^2 = u^2 + 2as

s = vt - \frac{1}{2}at^2

The Big Five (constant-acceleration equation strategy)

For constant acceleration in 1D, the “Big Five” idea is that the five kinematic quantities

v

v_0

a

\Delta x

t

are related by five equations, and each equation is missing one of the five quantities. A practical strategy is:

1) Identify which quantity is missing (neither given nor asked for).
2) Choose the equation that does not contain that quantity.

A common “Big Five” set (in terms of \Delta x = x - x_0) is:

Free fall as constant acceleration (and sign conventions)

Free fall is motion under the influence of gravity alone (ignoring air resistance). In free fall, an object is in a state of weightlessness in the sense of zero apparent weight.

Near Earth’s surface, the acceleration due to gravity has magnitude:

g \approx 9.8

and is approximately constant for all objects regardless of mass.

The sign of the acceleration depends on your coordinate choice:

• If you choose up as positive, then:

a = -g

• If you choose down as positive, then:

a = +g

In either case, the physics is the same if you stay consistent.

If you write free-fall equations with downward taken as positive (common in drop problems), you might see:

h = v_i t + \frac{1}{2}gt^2

v = v_i + gt

a = g

where h is downward displacement (or height fallen).

Terminal velocity is a real-world effect caused by air resistance: as speed increases, drag increases until drag balances weight, at which point acceleration becomes zero and the object falls at constant speed.

Example 1: interpreting signs in 1D

An object moves in 1D with:

v(t) = 4 - 3t

1) Acceleration:

a(t) = \frac{dv}{dt} = -3

2) When does it change direction?

4 - 3t = 0

t = \frac{4}{3}

Example 2: free fall with an upward-positive convention

A ball is thrown upward from height:

x_0 = 1.5

with initial velocity:

v_0 = 12

Take up as positive and use:

g = 9.8

Then:

a = -9.8

Time to reach the top occurs when:

v = 0

Using:

v = v_0 + at

gives:

0 = 12 - 9.8t

t = \frac{12}{9.8}

Position model:

x(t) = 1.5 + 12t + \frac{1}{2}(-9.8)t^2

A common mistake is to set acceleration to zero at the top because velocity is zero. Gravity does not “turn off” at the turning point.

Free fall sample problem (dropped rock)

A rock is dropped from an 80-meter cliff. How long does it take to reach the ground?

Choose down as positive so:

a = +g

Given:

\Delta y = 80

v_0 = 0

Use the constant-acceleration equation that does not involve final velocity (Big Five idea):

\Delta y = v_0 t + \frac{1}{2}at^2

So:

80 = \frac{1}{2}gt^2

t = \sqrt{\frac{160}{g}}

Using g = 9.8 gives approximately:

t \approx 4.04

Using g = 10 (a common approximation) gives exactly 4 s.

Exam Focus

• Typical question patterns:
  • Differentiate v(t) to get a(t), or differentiate x(t) twice.
  • Compute average acceleration from a change in velocity over a time interval.
  • Use sign analysis to determine when an object speeds up, slows down, or changes direction.
  • Apply constant-acceleration equations (including Big Five selection by “missing variable”) to straight-line motion and free fall.
• Common mistakes:
  • Thinking a > 0 means speeding up (ignoring the sign of v).
  • Using constant-acceleration formulas when acceleration is not constant.
  • Treating the top of a trajectory as a = 0.
  • Forgetting that free-fall sign depends on the chosen positive direction.

Connecting Motion with Integrals: From Acceleration to Velocity to Position

AP Physics C emphasizes that kinematics is a chain of relationships:

• acceleration is the derivative of velocity
• velocity is the derivative of position

You can reverse the chain using integrals, which is essential when acceleration is given as a function of time or when motion is described graphically.

The integral relationships

From:

a(t) = \frac{dv}{dt}

integrate from t_i to t_f:

\int_{t_i}^{t_f} a(t)dt = v(t_f) - v(t_i)

So change in velocity is the area under the acceleration-time curve.

From:

v(t) = \frac{dx}{dt}

integrate:

\int_{t_i}^{t_f} v(t)dt = x(t_f) - x(t_i)

So displacement is the signed area under the velocity-time curve.

Initial conditions and the constant of integration

Integration introduces an unknown constant that must be found using an initial condition.

If:

a(t) = 2t

then:

v(t) = t^2 + C

If:

v(0) = 3

then:

C = 3

and:

v(t) = t^2 + 3

Example 1: from acceleration to velocity to position

An object has:

a(t) = 6t

with:

v(0) = 2

x(0) = -1

Integrate acceleration:

v(t) = 3t^2 + C

Use v(0) = 2 to get:

v(t) = 3t^2 + 2

Integrate velocity:

x(t) = t^3 + 2t + K

Use x(0) = -1 to get:

x(t) = t^3 + 2t - 1

Example 2: displacement from a velocity-time graph (area)

If a velocity-time graph is a straight line from v = 2 at t = 0 to v = 8 at t = 3, then displacement is the trapezoid area:

\Delta x = \frac{1}{2}(v_i + v_f)\Delta t

\Delta x = \frac{1}{2}(2 + 8)(3) = 15

If the graph is below the time axis, the area is negative, indicating displacement in the negative direction.

Exam Focus

• Typical question patterns:
  • Given a(t), integrate to find v(t) and x(t) using initial conditions.
  • Use areas under a(t) and v(t) graphs to find changes in velocity or displacement.
  • Mix symbolic calculus with numeric evaluation at a particular time.
• Common mistakes:
  • Dropping the constant of integration.
  • Treating area under a velocity graph as distance instead of displacement.
  • Confusing “value of the function” with “area under the curve.”

Reading and Creating Kinematics Graphs (and What They Mean Physically)

Graphs are one of the fastest ways to communicate motion. Many AP questions embed reasoning in graphs without giving explicit equations, so you must be fluent with slope, curvature, and area.

Position-time graphs

• Slope of x(t) gives instantaneous velocity:

v(t) = \frac{dx}{dt}

• Concavity of x(t) relates to acceleration:

a(t) = \frac{d^2x}{dt^2}

Concave up means positive acceleration; concave down means negative acceleration. Direction comes from slope, not from whether the graph is above or below zero.

Velocity-time graphs

• Slope of v(t) gives acceleration:

a(t) = \frac{dv}{dt}

Because average acceleration is:

a_{avg} = \frac{\Delta v}{\Delta t}

the slope of a velocity-time graph over an interval gives the average acceleration for that interval.

• Area under v(t) gives displacement:

\Delta x = \int v(t)dt

Crossing v = 0 indicates a change in direction in 1D, not necessarily “rest for a while.”

Acceleration-time graphs

• Area under a(t) gives change in velocity:

\Delta v = \int a(t)dt

The height of an acceleration graph is not velocity; it is the rate of change of velocity.

Translating between graphs

• From x(t) to v(t): sketch slope versus time.
• From v(t) to a(t): sketch slope versus time.
• From a(t) to v(t): sketch accumulated area using the initial velocity.

Worked example 1: reasoning from a velocity-time graph (piecewise description)

A velocity-time graph is described as follows:

• At t = 0, v = 0.
• From t = 0 to t = 2, velocity increases steadily to 10.
• From t = 2 to t = 3, velocity decreases to 0.
• After t = 3, velocity becomes negative, reaching v = -5 at t = 3.5.
• From t = 3.5 onward, velocity remains constant at -5 (for example, through t = 6).

Acceleration from t = 0 to t = 2 is the slope:

a = \frac{10 - 0}{2 - 0} = 5

Acceleration from t = 2 to t = 3.5 is the slope between the points:

a = \frac{-5 - 10}{3.5 - 2} = -10

Acceleration from t = 3.5 to t = 6 is zero because the graph is horizontal.

Worked example 2: consistent graph reasoning (slope and area)

An object’s velocity graph is a straight line from v = 6 at t = 0 to v = -2 at t = 4.

1) Acceleration (slope):

a = \frac{-2 - 6}{4 - 0} = -2

2) Time when it changes direction (when v = 0). The drop from 6 to 0 is 6 units out of a total drop of 8 units over 4 s:

t = 4\left(\frac{6}{8}\right) = 3

3) Displacement from t = 0 to t = 4 is the signed area under v(t).

From 0 to 3 (velocities 6 to 0):

\Delta x_{0\to3} = \frac{1}{2}(6 + 0)(3) = 9

From 3 to 4 (velocities 0 to -2):

\Delta x_{3\to4} = \frac{1}{2}(0 - 2)(1) = -1

Total:

\Delta x = 8

Exam Focus

• Typical question patterns:
  • Use slopes of graphs to identify velocity or acceleration.
  • Use areas under graphs to compute displacement or change in velocity.
  • Interpret a piecewise velocity graph to find accelerations over different intervals.
  • Choose which graph matches a motion description (qualitative translation).
• Common mistakes:
  • Reading the value of x(t) as velocity or the value of v(t) as acceleration.
  • Forgetting that area can be negative when the graph is below the axis.
  • Assuming “crossing zero” implies staying at rest for a time interval.

Motion in Two Dimensions: Vectors, Components, and Parametric Models

Two-dimensional kinematics uses the same ideas as 1D kinematics, but applied to vectors. The key skill is thinking in components: in many problems, perpendicular components evolve independently.

Vector quantities (what they are and how to work with them)

A vector quantity has both magnitude and direction. It can be represented by a directed line segment (arrow): the arrow length represents magnitude and the arrow direction represents the vector direction. Vectors are often written in boldface in print (for example v for velocity or F for force).

A unit vector is a vector of magnitude 1.

Addition of vectors can be done using the head-to-tail method or the parallelogram method. The resultant vector runs from the tail of the first vector to the head of the last vector.

Subtraction of vectors is performed by adding the negative vector. The negative of a vector has the same magnitude but points in the opposite direction.

Scalar multiplication multiplies a vector’s magnitude by the scalar. The direction stays the same unless the scalar is negative, which reverses the direction.

Position, velocity, and acceleration vectors

In 2D:

\vec r(t) = x(t)\hat i + y(t)\hat j

Velocity is the derivative:

\vec v(t) = \frac{d\vec r}{dt} = \frac{dx}{dt}\hat i + \frac{dy}{dt}\hat j

Acceleration is the derivative of velocity:

\vec a(t) = \frac{d\vec v}{dt} = \frac{d^2x}{dt^2}\hat i + \frac{d^2y}{dt^2}\hat j

So the component relationships are:

v_x(t) = \frac{dx}{dt}

v_y(t) = \frac{dy}{dt}

a_x(t) = \frac{dv_x}{dt}

a_y(t) = \frac{dv_y}{dt}

Vector magnitudes

Speed is the magnitude of velocity:

|\vec v| = \sqrt{v_x^2 + v_y^2}

Acceleration magnitude is:

|\vec a| = \sqrt{a_x^2 + a_y^2}

Even if v_x is constant (as in ideal projectile motion), |\vec v| usually changes because v_y changes.

Components from magnitude and angle

If an object is launched with speed v_0 at angle \theta above the horizontal:

v_{0x} = v_0\cos\theta

v_{0y} = v_0\sin\theta

A quick self-check: if \theta = 0, you should get v_{0x} = v_0 and v_{0y} = 0.

Example: building a parametric model

A particle’s position is:

\vec r(t) = (2t^2)\hat i + (5 - 3t)\hat j

Differentiate:

\vec v(t) = (4t)\hat i + (-3)\hat j

Differentiate again:

\vec a(t) = (4)\hat i + (0)\hat j

Interpretation: the particle has constant downward velocity component and a constant rightward acceleration.

Exam Focus

• Typical question patterns:
  • Differentiate component functions to obtain \vec v(t) and \vec a(t).
  • Compute speed from components at a given time.
  • Decompose an initial velocity into components using an angle.
  • Add and subtract vectors correctly (including “subtract by adding the negative”).
• Common mistakes:
  • Treating vectors like scalars (mixing up magnitude with components).
  • Forgetting that \vec v and \vec a can point in different directions.
  • Swapping sine and cosine in component decomposition.
  • Adding magnitudes when the situation requires vector addition.

Projectile Motion (No Air Resistance): The Classic 2D Kinematics Model

Projectile motion is motion where the only acceleration is gravity (air resistance neglected). It is heavily tested because it combines components, constant acceleration, and careful interpretation of what changes and what does not.

Core physical model

For ideal projectile motion:

a_x = 0

a_y = -g

So:

v_x(t) = v_{0x}

v_y(t) = v_{0y} - gt

x(t) = x_0 + v_{0x}t

y(t) = y_0 + v_{0y}t - \frac{1}{2}gt^2

The vertical position is quadratic in time, producing a parabolic path.

Why the components are independent

Gravity acts vertically, so horizontal motion does not “know” about gravity in this model. A common misconception is that greater horizontal speed keeps the projectile in the air longer. In ideal projectile motion, time in the air depends only on vertical motion.

Symmetry ideas (same launch and landing height)

If the projectile lands at the same height it was launched:

• Time up equals time down.
• The landing vertical velocity has the same magnitude as the launch vertical velocity, but opposite sign.

Example 1: time of flight and range (same launch and landing height)

A projectile is launched from ground with speed v_0 at angle \theta.

Time of flight comes from vertical motion with y = 0 at launch and landing:

0 = v_{0y}t - \frac{1}{2}gt^2

Nonzero solution:

t = \frac{2v_{0y}}{g}

Using v_{0y} = v_0\sin\theta:

t = \frac{2v_0\sin\theta}{g}

Range is:

R = v_{0x}t

Using v_{0x} = v_0\cos\theta:

R = \frac{v_0^2\sin(2\theta)}{g}

The maximum range occurs when:

\theta = 45

Projectile sample problem 1 (numerical)

An object is projected with launch angle 30 degrees and initial speed 60 m/s. How long is it in the air, and how far does it travel horizontally before returning to its original height?

Vertical component:

v_{0y} = 60\sin 30 = 30

Time to the top (where v_y = 0):

t_{up} = \frac{v_{0y}}{g}

Using the common approximation g = 10 gives:

t_{up} = 3

Total flight time (symmetry):

T = 2t_{up} = 6

Horizontal component:

v_{0x} = 60\cos 30 = 30\sqrt{3}

Horizontal distance (range):

R = v_{0x}T = (30\sqrt{3})(6) = 180\sqrt{3}

Projectile sample problem 2 (time to top and max height)

An object is projected with launch angle 30 degrees and initial speed 40 m/s. How long to reach the top, and how high is it?

Vertical component:

v_{0y} = 40\sin 30 = 20

At the top, v_y = 0. Using g = 10:

0 = 20 - 10t

t = 2

Maximum height above launch point:

\Delta y = v_{0y}t - \frac{1}{2}gt^2 = 20(2) - \frac{1}{2}(10)(2^2) = 20

Example 2: projectile launched from a height

A ball is kicked from height y_0 with initial vertical component v_{0y} and lands at ground level y = 0. Use:

0 = y_0 + v_{0y}t - \frac{1}{2}gt^2

Solve the quadratic for t and choose the positive root. Then horizontal distance is:

\Delta x = v_{0x}t

A common mistake is using the same-height time-of-flight formula when launch and landing heights differ.

Example 3: speed at a point

Even though v_x is constant, the speed generally changes:

|\vec v(t)| = \sqrt{v_{0x}^2 + (v_{0y} - gt)^2}

Real-world connection

Real projectiles experience air resistance, so the path is not a perfect parabola. The ideal model is still valuable because it isolates gravity’s role and builds component-based reasoning that many more advanced models build on.

Exam Focus

• Typical question patterns:
  • Use separate x and y equations to solve for time, range, maximum height, or landing speed.
  • Identify which quantities remain constant (typically v_x) and which change (typically v_y).
  • Solve for the launch angle or initial speed given range and height conditions.
  • Use symmetry only when launch and landing heights match.
• Common mistakes:
  • Assuming time of flight depends on horizontal velocity.
  • Using same-height symmetry formulas when launch and landing heights differ.
  • Plugging speed directly into 1D formulas without separating into components.

Relative Motion and Changing Reference Frames (Galilean Transformations)

Relative motion problems compare measurements made in different frames (for example, a person walking on a moving train or a boat in a river). In AP Physics C kinematics, these are treated with Galilean relativity: time is the same in all frames, and velocities add linearly.

Relative position and relative velocity

In 1D:

x_{A/B} = x_A - x_B

Differentiate to get relative velocity:

v_{A/B} = v_A - v_B

In vector form:

\vec r_{A/B} = \vec r_A - \vec r_B

\vec v_{A/B} = \vec v_A - \vec v_B

Velocity addition (frame transformations)

If frame S' moves with velocity \vec u relative to frame S, and the object’s velocity in S is \vec v, then:

\vec v' = \vec v - \vec u

Equivalently:

\vec v = \vec v' + \vec u

A reliable way to avoid sign mistakes is to translate the relationship into a sentence before writing symbols.

Example 1: two cars passing

Car A moves east at 25. Car B moves east at 18. Taking east as positive:

v_{A/B} = 25 - 18 = 7

So A moves east relative to B at 7.

And:

v_{B/A} = 18 - 25 = -7

The negative sign means B moves west relative to A.

Example 2: river boat (2D relative velocity)

If the boat’s velocity relative to water is \vec v_{B/W} and the water’s velocity relative to ground is \vec v_{W/G}, then:

\vec v_{B/G} = \vec v_{B/W} + \vec v_{W/G}

A common pitfall is adding speeds (magnitudes) instead of velocity vectors. Direction matters, and the boat often must aim upstream to travel straight across relative to the ground.

Exam Focus

• Typical question patterns:
  • Compute relative velocity between two objects moving along a line.
  • Use vector addition for boat-in-river or plane-in-wind style problems.
  • Interpret negative relative velocity as “moving opposite direction in that frame.”
• Common mistakes:
  • Adding magnitudes instead of vectors.
  • Flipping the order in v_{A/B} = v_A - v_B.
  • Forgetting to specify a positive direction and then misreading signs.

Multi-Stage Motion and Piecewise Models (When the Rules Change Mid-Problem)

Many motions happen in stages: accelerate, cruise, then brake. The correct method is to build a separate model for each stage and then stitch the stages together with boundary conditions.

Piecewise functions and continuity

If acceleration changes at time t_1, you may define stages such as:

• Stage 1 for 0 \le t \le t_1 with acceleration a_1(t)
• Stage 2 for t_1 \le t \le t_2 with acceleration a_2(t)

At the boundary time t_1, position should match:

x_1(t_1) = x_2(t_1)

Velocity should also match if there is no sudden jump:

v_1(t_1) = v_2(t_1)

Example: accelerate then coast

A car starts at:

x_0 = 0

v_0 = 0

It accelerates at constant:

a = 2

for 5 s, then continues at constant velocity for 10 more seconds.

Stage 1 (0 to 5):

v(5) = 0 + (2)(5) = 10

x(5) = 0 + 0\cdot 5 + \frac{1}{2}(2)(5^2) = 25

Stage 2 (5 to 15): velocity is constant at 10, so:

\Delta x_2 = v\Delta t = 10\cdot 10 = 100

Total displacement:

x(15) = 25 + 100 = 125

A common mistake is to keep using the constant-acceleration position formula in stage 2 even though acceleration is zero there.

Exam Focus

• Typical question patterns:
  • Piecewise motion described in words or graphs, requiring separate calculations and then summing intervals.
  • Use boundary conditions to link stages (especially velocity at the transition).
  • Determine total displacement or total time given different acceleration regimes.
• Common mistakes:
  • Using a single kinematic equation across a time interval where acceleration changes.
  • Forgetting to carry final values from stage 1 as initial values for stage 2.
  • Adding velocities or accelerations across stages instead of adding displacements or times appropriately.