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1.1 Scalars and Vectors in One Dimension

A scalar answers ‘how much?’ A vector answers ‘how much and in which direction?’

Quantity	Type	What must be specified
Distance	Scalar	Magnitude only
Speed	Scalar	Magnitude only
Position	Vector	Location relative to an origin, including direction
Displacement	Vector	Change in position, including direction
Velocity	Vector	Rate of change of position, including direction
Acceleration	Vector	Rate of change of velocity, including direction

In one dimension, a sign can encode direction only after a positive axis has been chosen. A negative component means ‘opposite the positive direction’—not ‘slower’ or ‘smaller.’

Set the direction convention

Choose one positive direction. Write vectors in that direction as positive and vectors in the opposite direction as negative, then add their one-dimensional components.

Add the signed components

$\Delta x_{\text{net}}=(+5\,\mathrm{m})+(-2\,\mathrm{m})=+3\,\mathrm{m}$

For this hypothetical motion, take right as positive: a 5 m displacement to the right is $+5\,\mathrm{m}$ and a 2 m displacement to the left is $-2\,\mathrm{m}$ . The resultant is $+3\,\mathrm{m}$ , so its magnitude is 3 m and its direction is right.

Interpret the sign

The sign of the answer depends on the chosen axis; the physical resultant does not. Choosing left as positive would give $-3\,\mathrm{m}$ , which still describes 3 m to the right.

1.2 Displacement, Velocity, and Acceleration

Model the object

In the object model, treat the object as a single point. Its size, shape and internal details are ignored so its position can be tracked on a chosen axis.

Compare the endpoints

$\Delta x=x-x_0$

Hypothetical example: choose right as positive. An object moves from $x_0=+2\,\mathrm{m}$ to $x=-3\,\mathrm{m}$ . Then $\Delta x=(-3\,\mathrm{m})-(+2\,\mathrm{m})=-5\,\mathrm{m}$ . The magnitude is 5 m and the negative sign means the displacement is to the left.

Do not confuse path and change

Displacement depends only on the initial and final positions; distance traveled depends on the whole path. Returning to the starting point gives zero displacement even when the distance traveled is not zero.

Match each rate to its change

Average quantity	Change in the numerator	Relationship	SI unit
Velocity	Displacement, $\Delta x$	$\bar v=\dfrac{\Delta x}{\Delta t}$	$\mathrm{m\,s^{-1}}$
Acceleration	Velocity, $\Delta v=v-v_0$	$\bar a=\dfrac{\Delta v}{\Delta t}$	$\mathrm{m\,s^{-2}}$

Substitute with signs and units

$\bar v=\frac{+12\,\mathrm{m}}{4\,\mathrm{s}}=+3\,\mathrm{m\,s^{-1}}$

In the same hypothetical interval, suppose velocity changes from $v_0=+4\,\mathrm{m\,s^{-1}}$ to $v=-4\,\mathrm{m\,s^{-1}}$ . Then $\Delta v=-8\,\mathrm{m\,s^{-1}}$ and $\bar a=(-8\,\mathrm{m\,s^{-1}})/(4\,\mathrm{s})=-2\,\mathrm{m\,s^{-2}}$ . The negative sign means the average acceleration points in the negative direction.

Interpret acceleration carefully

Acceleration occurs when velocity changes in magnitude and/or direction. Negative acceleration does not automatically mean slowing down. As the time interval becomes very small, an average value approaches the corresponding instantaneous value.

1.3 Representing Motion

One motion, several representations

A motion diagram, graph, equation, figure or narrative can describe the same motion. A valid translation preserves the signs, changes and time interval represented by position $x$ , velocity $v_x$ and acceleration $a_x$ .

Choose an equation by what is known

Use when the unknown is…	Constant-acceleration relationship
Final velocity after time $t$	$v_x=v_{x0}+a_xt$
Position after time $t$	$x=x_0+v_{x0}t+\tfrac12a_xt^2$
Velocity after a displacement, with no time needed	$v_x^2=v_{x0}^2+2a_x(x-x_0)$

Read slopes and signed areas

Representation	Read this feature	It gives…
Position–time graph	Tangent slope	Instantaneous velocity
Velocity–time graph	Tangent slope	Instantaneous acceleration
Velocity–time graph	Signed area over an interval	Displacement
Acceleration–time graph	Signed area over an interval	Change in velocity

Check that the representations agree

Hypothetical example: a point object starts at $x_0=1\,\mathrm{m}$ with $v_{x0}=0$ and constant $a_x=+2\,\mathrm{m\,s^{-2}}$ for $t=3\,\mathrm{s}$ . Then $v_x=0+(2)(3)=+6\,\mathrm{m\,s^{-1}}$ and $x=1+0+\tfrac12(2)(3^2)=10\,\mathrm{m}$ . Its displacement is $+9\,\mathrm{m}$ , equal to the area under its velocity–time graph.

Conditions and boundaries

The three equations require constant acceleration. AP Physics 1 treats nonuniform acceleration quantitatively only at a qualitative graph level. Near Earth, gravitational acceleration is downward with magnitude $g\approx10\,\mathrm{m\,s^{-2}}$ ; its component is negative if upward is chosen positive.

1.4 Reference Frames and Relative Motion

Name the frame

For one-dimensional motion, an observer measures the object’s velocity relative to the observer’s own frame. Choose a positive direction and keep every velocity signed.

Subtract the observer velocity

$v_{\text{object/observer}}=v_{\text{object/ground}}-v_{\text{observer/ground}}$

Hypothetical example: east is positive. A passenger walks east at $+2\,\mathrm{m\,s^{-1}}$ relative to a train moving east at $+10\,\mathrm{m\,s^{-1}}$ relative to the ground. The passenger’s ground velocity is $+12\,\mathrm{m\,s^{-1}}$ ; equivalently, a ground observer moving at 0 measures 12 m/s east, while a train observer measures $12-10=+2\,\mathrm{m\,s^{-1}}$ .

What changes—and what does not

Velocity can differ between inertial frames, but acceleration is the same in all inertial frames. AP Physics 1 restricts relative-velocity vector addition or subtraction to one dimension and assumes the stated frame is inertial unless told otherwise.

Define the observer’s coordinates

A reference frame is the coordinate system and clock attached to an observer. Identify the observer, origin, positive axis and time reference before describing position or velocity.

Compare the same event

Same event	Ground frame	Train frame
Train moving east at 10 m/s	Train velocity is $+10\,\mathrm{m\,s^{-1}}$	Train velocity is $0$
Seat fixed inside the train	Seat changes ground position	Seat position is constant

Changing frames can change the measured position and velocity—both magnitude and direction—but does not change the underlying event. State conclusions as ‘relative to the ground’ or ‘relative to the train.’

Use the inertial-frame assumption

Different frame-dependent values are not contradictory. Unless a problem says otherwise, AP Physics 1 lets you assume the reference frame is inertial: it is not accelerating or rotating relative to another inertial frame.

1.5 Vectors and Motion in Two Dimensions

Choose perpendicular axes

Choose perpendicular $x$ and $y$ axes. A vector $\vec A$ can be replaced by components $A_x$ and $A_y$ whose vector sum is the original vector.

Resolve and reconstruct

If $\theta$ is measured from the +x-axis	Relationship
Adjacent component	$A_x=A\cos\theta$
Opposite component	$A_y=A\sin\theta$
Reconstruct magnitude	$A=\sqrt{A_x^2+A_y^2}$
Reconstruct direction	$\tan\theta=A_y/A_x$

Calculate both components

Hypothetical example: a $10\,\mathrm{N}$ vector points $30^\circ$ above +x. Then $A_x=(10)\cos30^\circ=8.66\,\mathrm{N}$ and $A_y=(10)\sin30^\circ=5.00\,\mathrm{N}$ . Check: $\sqrt{8.66^2+5.00^2}=10.0\,\mathrm{N}$ .

Keep coordinate signs

The trigonometric magnitudes do not replace the coordinate signs. Determine each sign from the vector’s direction relative to the chosen axes. Components are simultaneous parts of one vector—not two vectors applied one after the other.

Separate the components

Resolve position, velocity and acceleration into perpendicular components. Analyze the $x$ and $y$ motions separately with one-dimensional kinematics, then combine them at the same time $t$ .

Compare horizontal and vertical motion

Projectile component	Acceleration	Consequence
Horizontal $x$	$a_x=0$	$v_x$ is constant and $\Delta x=v_xt$
Vertical $y$	$a_y=-g$ if up is positive	$v_y$ changes at a constant rate

Use one shared time

Hypothetical example: a projectile launches horizontally at $v_x=6\,\mathrm{m\,s^{-1}}$ from $20\,\mathrm{m}$ above the ground. With up positive and $g=10\,\mathrm{m\,s^{-2}}$ , $-20=\tfrac12(-10)t^2$ , so $t=2\,\mathrm{s}$ . During that same time, $\Delta x=(6)(2)=12\,\mathrm{m}$ .

Recombine the story

Zero horizontal acceleration does not mean zero horizontal velocity. The components evolve independently under their own accelerations, but they are synchronized by the same time interval and together form one curved trajectory.

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