Unit 1: 1D and 2D Kinematics

Scalars and Vectors

A scalar has magnitude only (examples: speed, distance, time). A vector is a quantity that has magnitude and direction (examples: displacement, velocity, acceleration).

Graphically, a vector is drawn as an arrow. The length of the arrow represents the magnitude, and the direction of the arrow represents the direction of the quantity. You can slide a vector around without changing it as long as its length and direction stay the same. For example, a displacement of $5\ \text{m}$ east is the same vector no matter where you draw the arrow on the page.

If a vector $\vec A$ makes an angle $\theta$ above the positive $x$ -axis, its components are

A_x=A\cos\theta, \qquad A_y=A\sin\theta.

This comes straight from right-triangle trig: the horizontal component is adjacent to $\theta$ and the vertical component is opposite $\theta$ . If the vector points left or down, the corresponding component should be negative. The trig formulas give the correct signs automatically if $\theta$ is measured from the positive $x$ -axis in standard position. $A_y$ is drawn twice to better illustrate the Pythagorean relation.

If instead you know the components, then the magnitude and direction come from

A=\sqrt{A_x^2+A_y^2}, \qquad \tan\theta=\frac{A_y}{A_x}.

Be careful with the angle formula. The value of $\tan^{-1}(A_y/A_x)$ may point to the wrong quadrant if $A_x$ is negative, so always check the signs of the components.

Example. A velocity vector has magnitude $24\ \text{m/s}$ and points $35^\circ$ north of west. Find its $x$ - and $y$ -components if east is positive $x$ and north is positive $y$ .

The phrase “north of west” means the vector starts by pointing west, then rotates $35^\circ$ toward north. So the $x$ -component is negative and the $y$ -component is positive.

The horizontal component uses cosine because it is adjacent to the $35^\circ$ angle:

v_x=-24\cos35^\circ\approx -19.7\ \text{m/s}.

The vertical component uses sine because it is opposite the angle:

v_y=24\sin35^\circ\approx 13.8\ \text{m/s}.

Thus

\vec v\approx (-19.7\hat{i}+13.8\hat{j})\ \text{m/s}.

Unit vectors

Using unit vectors, a two-dimensional vector can be written as

\vec A=A_x\hat{i}+A_y\hat{j},

where $\hat{i}$ points in the positive $x$ direction and $\hat{j}$ points in the positive $y$ direction. In three dimensions, we add $\hat{k}$ for the positive $z$ direction:

\vec A=A_x\hat{i}+A_y\hat{j}+A_z\hat{k}.

Example. A particle has displacement $\Delta \vec r=(6\hat{i}-8\hat{j})\ \text{m}$ . Find its displacement magnitude and the unit vector in the direction of the displacement.

The magnitude comes from the Pythagorean theorem:

\lvert \Delta \vec r\rvert=\sqrt{6^2+(-8)^2}=10\ \text{m}.

A unit vector keeps the direction but has magnitude $1$ . Divide the vector by its magnitude:

\hat{r}=\frac{\Delta \vec r}{\lvert \Delta \vec r\rvert} =\frac{6\hat{i}-8\hat{j}}{10}.

\hat{r}=0.6\hat{i}-0.8\hat{j}.

This answer has no units because unit vectors describe direction only.

Vector addition/subtraction

Vector addition can be done graphically by placing arrows head-to-tail. The resultant points from the tail of the first vector to the head of the last vector. Note that the order of addition doesn’t matter and would converge at a point regardless of the addition order.

You can also add vectors component wise:

\vec A+\vec B=(A_x+B_x)\hat{i}+(A_y+B_y)\hat{j}.

Subtraction means adding the opposite vector:

\vec A-\vec B=\vec A+(-\vec B)=(A_x-B_x)\hat{i}+(A_y-B_y)\hat{j}.

Note that for subtraction, order does matter, since subtraction is not commutative.

In one dimension, a sign attached to a component is enough to encode direction along an axis. For example, if right is positive, then $v=-3\ \text{m/s}$ means the object moves left at $3\ \text{m/s}$ . Before solving any kinematics problem, choose an origin and positive direction for each axis, then keep that convention consistent.

Example. A vector has magnitude $10\ \text{m}$ and points $30^\circ$ above the positive $x$ -axis. Another displacement is $\vec B=(-2\hat{i}+4\hat{j})\ \text{m}$ . Find $\vec A+\vec B$ and its magnitude.

First find the components of $\vec A$ :

A_x=10\cos30^\circ=5\sqrt{3}, \qquad A_y=10\sin30^\circ=5.

\vec A=(5\sqrt{3}\hat{i}+5\hat{j})\ \text{m}.

Add components:

\vec A+\vec B=(5\sqrt{3}-2)\hat{i}+(5+4)\hat{j}.

Thus

\vec A+\vec B=((5\sqrt{3}-2)\hat{i}+9\hat{j})\ \text{m}.

The magnitude is

\lvert \vec A+\vec B\rvert =\sqrt{(5\sqrt{3}-2)^2+9^2} \approx 11.2\ \text{m}.

Variables of motion

Displacement $\Delta x$ is the change in position, regardless of the path taken, and can be negative. Distance is the length of that path and is always nonnegative. Average velocity over an interval is defined as

\bar{v} = \frac{\Delta x}{\Delta t}.

Instantaneous velocity is the time derivative of position when position is given as a function $x(t)$ :

v = \frac{dx}{dt}.

Speed is the magnitude of velocity, $\lvert v \rvert$ , in one dimension.

Average acceleration is defined as:

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

Instantaneous acceleration is the derivative of velocity with respect to time, equivalently the second derivative of position:

a = \frac{dv}{dt} = \frac{d^2x}{dt^2}.

If you know $a(t)$ , the change in velocity between times $t_1$ and $t_2$ follows from integration:

v(t_2) - v(t_1) = \int_{t_1}^{t_2} a(t)\, dt,

and similarly position from velocity.

Always remember the difference between average and instantaneous quantities. Average velocity over an interval is the single constant velocity that would produce the same displacement in the same time. It depends only on the endpoints, $\bar{v}=\Delta x/\Delta t$ .

Instantaneous velocity is the limit of that ratio as the interval shrinks to zero, $v=dx/dt$ . The two agree only when velocity is constant, or, for the special case of constant acceleration, the average velocity happens to equal the midpoint value $\frac{1}{2}(v_0+v_f)$ .

Example. A particle starts at $x_0 = 0$ with velocity $v_0 = 2\ \text{m/s}$ and experiences acceleration $a(t) = 6t\ \text{m/s}^2$ . Find its velocity and position as functions of time, and its position at $t = 3\ \text{s}$ .

Integrate acceleration to get velocity:

v(t) = v_0 + \int_0^t 6t'\, dt' = 2 + 3t^2.

Integrate velocity to get position:

x(t) = x_0 + \int_0^t (2 + 3t'^2)\, dt' = 2t + t^3.

At $t = 3\ \text{s}$ ,

x(3) = 2(3) + (3)^3 = 6 + 27 = 33\ \text{m}.

Constant acceleration in one dimension

Many problems use constant acceleration $a$ (free fall near Earth’s surface is a common case with $a = -g$ or $a = +g$ depending on axis choice). The following equations (known as the Big Five) are very useful for these types of problems, since they only require 4 out of the 5 useful variables ( $a$ , $v_f$ , $v_0$ , $\Delta x$ , $t$ ). By convention, we set $t_0 = 0$ , with $a$ being acceleration, $v_0$ being initial velocity, $v_f$ being final velocity, $\Delta x$ being displacement, and $t$ being time:

These are algebraic consequences of $a = dv/dt$ constant and $v = dx/dt$ .

Proof (The Big Five). Start with constant acceleration:

a=\frac{dv}{dt}.

Since $a$ is constant, integrate from $0$ to $t$ :

\int_{v_0}^{v}dv=\int_0^t a\,dt.

This gives

v-v_0=at,

v=v_0+at.

That proves equation 1.

Since velocity is the derivative of position,

v=\frac{dx}{dt}.

Using $v(t)=v_0+at$ ,

\Delta x=\int_0^t v(t)\,dt=\int_0^t (v_0+at)\,dt.

Therefore

\Delta x=v_0t+\frac{1}{2}at^2.

That proves equation 2.

Solve equation 1 for the initial velocity:

v_0=v-at.

Substitute into equation 2:

\Delta x=(v-at)t+\frac{1}{2}at^2.

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

That proves equation 3.

To eliminate time, use the chain rule:

a=\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}.

Then

a\,dx=v\,dv.

Integrate from $x_0$ to $x$ and from $v_0$ to $v$ :

\int_{x_0}^{x}a\,dx=\int_{v_0}^{v}v\,dv.

a\Delta x=\frac{1}{2}(v^2-v_0^2),

which rearranges to

v^2=v_0^2+2a\Delta x.

That proves equation 4.

Finally, for constant acceleration, the velocity-time graph is a straight line, so displacement is the area under that graph: a trapezoid with bases $v_0$ and $v$ and width $t$ . Thus

\Delta x=\frac{v_0+v}{2}t.

That proves equation 5.

Example. A car traveling at $v_0 = 25\ \text{m/s}$ brakes with constant deceleration and comes to rest in $40\ \text{m}$ . Find the acceleration and the time it takes to stop.

The final velocity is $v = 0$ , and we know $v_0$ and $\Delta x$ but not $t$ , so use equation 4 (missing $t$ ):

v^2 = v_0^2 + 2a\Delta x \;\Rightarrow\; 0 = (25)^2 + 2a(40).

Solve:

a = -\frac{625}{80} = -7.8\ \text{m/s}^2.

The negative sign means the acceleration opposes the motion, as expected for braking. For the time, use equation 1:

v = v_0 + at \;\Rightarrow\; 0 = 25 + (-7.8)t,

t = \frac{25}{7.8} \approx 3.2\ \text{s}.

A quick check with equation 5: $\Delta x = \frac{v_0 + v}{2}t = \frac{25 + 0}{2}(3.2) \approx 40\ \text{m}$ , which matches.

Motion graphs

Graphs of position, velocity, and acceleration versus time are linked by the same calculus that links the quantities themselves.

Slopes go down the list. The slope of an $x$ - $t$ graph at an instant is the velocity, $v = dx/dt$ . The slope of a $v$ - $t$ graph is the acceleration, $a = dv/dt$ . A curving $x$ - $t$ graph therefore means nonzero acceleration, and a straight $x$ - $t$ line means constant velocity.
Areas go up the list. The signed area under a $v$ - $t$ graph between two times is the displacement, $\Delta x = \int v\, dt$ . The signed area under an $a$ - $t$ graph is the change in velocity, $\Delta v = \int a\, dt$ . In both cases, count areas below the time axis as negative.

A maximum or minimum of $x(t)$ occurs where $v = 0$ (the slope is momentarily flat); the object is speeding up when $v$ and $a$ have the same sign and slowing down when they have opposite signs; and a horizontal $v$ - $t$ line means zero acceleration even if the velocity itself is large.

If you are struggling to draw a motion graph, always consider the important components of the graph, like turning points, end behavior, or concavity. Most problems do not require a very detailed/completely accurate motion graph.

Example. A cart moves along a line. Its velocity is $+4\ \text{m/s}$ and constant for the first $3\ \text{s}$ , then ramps linearly down to $-2\ \text{m/s}$ over the next $2\ \text{s}$ . Find the acceleration during each phase, the total displacement, and the distance traveled.

During the first phase the $v$ - $t$ graph is flat, so $a_1 = 0$ . During the second phase,

a_2 = \frac{\Delta v}{\Delta t} = \frac{(-2) - (+4)}{2} = -3\ \text{m/s}^2.

Displacement is the signed area under the $v$ - $t$ graph. The first phase is a rectangle:

\Delta x_1 = (4)(3) = 12\ \text{m}.

In the second phase the velocity crosses zero. It hits $v=0$ after $\Delta v / a_2 = (-4)/(-3) = \frac{4}{3}\ \text{s}$ , moving forward, then reverses. The signed area of the triangular region is

\Delta x_2 = \tfrac{1}{2}(4)\left(\tfrac{4}{3}\right) + \tfrac{1}{2}(-2)\left(2-\tfrac{4}{3}\right) = \tfrac{8}{3} - \tfrac{2}{3} = 2\ \text{m}.

So total displacement is $12 + 2 = 14\ \text{m}$ . Distance traveled adds the magnitudes of the forward and backward pieces: $12 + \tfrac{8}{3} + \tfrac{2}{3} = 12 + \tfrac{10}{3} \approx 15.3\ \text{m}$ . Displacement and distance differ because the cart briefly reverses.

Non-constant acceleration

The Big Five only works in constant acceleration, so when you have non-constant $s$ you often fall back on the defining derivatives and choose your integration variable based on what $a$ depends on:

Problem-solving strategy

$a$ depends on time, $a = a(t)$ . Integrate directly: $v = v_0 + \int a(t)\, dt$ , then $x = x_0 + \int v(t)\, dt$ .
$a$ depends on position, $a = a(x)$ . Time is not very useful here, so eliminate it with the chain rule. Writing $a = dv/dt = (dv/dx)(dx/dt) = v\,dv/dx$ gives the separable relation $a(x)\, dx = v\, dv$ . Integrating both sides connects speed to position without ever solving for time, much like how equation 4 was derived.
$a$ depends on velocity, $a = a(v)$ . Separate variables in $dv/dt = a(v)$ to get $dt = dv/a(v)$ , or use $v\,dv/dx = a(v)$ if you want $v$ as a function of $x$ .

Always remember to use the chain rule when you want a variable that is not in the original expression.

Example. A particle moves along the $x$ -axis with acceleration $a(t)=6t$ . At $t=0$ , it has $v_0=2.0\ \text{m/s}$ and $x_0=1.0\ \text{m}$ . Find $v(t)$ and $x(t)$ .

Since acceleration is a function of time, integrate directly:

v(t)=v_0+\int_0^t 6t'\,dt' =2.0+3t^2.

Then integrate velocity:

x(t)=x_0+\int_0^t \left(2.0+3t'^2\right)\,dt' =1.0+2.0t+t^3.

The Big Five do not apply because acceleration is not constant. The definitions $a=dv/dt$ and $v=dx/dt$ are the safer starting point.

Drag acceleration

Drag is a form nonconstant acceleration and is the most comon type of non-constant acceleration found in AP Physics. Here, we model its effect directly as an acceleration, even though it is technically a force. In any fluid (including gases like air), a moving object will collide with particles in the fluid, causing a friction force that opposes the motion of the object. For downward motion through still air, the drag acceleration points upward, opposing gravity.

Two common drag models are:

Key equations

Model	Drag acceleration magnitude	When it is used
Linear drag	$a_d=bv$	slower motion through a fluid
Quadratic drag	$a_d=cv^2$	faster motion through air, more realistic for many projectiles

We use linear drag here because its differential equation is simpler to solve than the quadratic model. If downward is positive for a falling object, then gravity is positive and drag is negative:

a=\frac{dv}{dt}=g-bv.

At first, $v=0$ , so drag is zero and the acceleration is $g$ . As the object speeds up, $bv$ grows, so the acceleration decreases. Eventually the total acceleration becomes zero, and the object reaches terminal velocity:

g-bv_T=0 \quad\Rightarrow\quad v_T=\frac{g}{b}.

For linear drag, the differential equation can actually be solved.

Example. A ball is dropped from rest through air with linear drag acceleration $a_d=bv$ upward. Take downward as positive. Find $v(t)$ , then find the terminal velocity in two ways.

Since downward is positive, gravitational acceleration is $+g$ and drag acceleration is $-bv$ . Therefore

\frac{dv}{dt}=g-bv.

We want velocity as a function of time, so separate variables:

\frac{dv}{g-bv}=dt.

Because the ball is dropped from rest, $v(0)=0$ . Integrate from the initial velocity $0$ to the velocity at time $t$ :

\int_0^{v(t)} \frac{1}{g-bv}\,dv=\int_0^t dt.

For the left side, use the substitution

u=g-bv, \qquad du=-b\,dv.

When $v=0$ , $u=g$ . When $v=v(t)$ , $u=g-bv(t)$ . Therefore

\int_0^{v(t)} \frac{1}{g-bv}\,dv =-\frac{1}{b}\int_g^{g-bv(t)}\frac{1}{u}\,du.

Evaluate the integral:

-\frac{1}{b}\int_g^{g-bv(t)}\frac{1}{u}\,du =-\frac{1}{b}\left[\ln\lvert u\rvert\right]_g^{g-bv(t)}.

-\frac{1}{b}\left(\ln\lvert g-bv(t)\rvert-\ln g\right)=t.

Combine the logarithms:

-\frac{1}{b}\ln\left\lvert \frac{g-bv(t)}{g}\right\rvert=t.

Multiply by $-b$ :

\ln\left\lvert \frac{g-bv(t)}{g}\right\rvert=-bt.

Since the ball starts from rest and approaches terminal velocity from below, $0\le v(t)<\frac{g}{b}$ while it is speeding up. Thus $g-bv(t)>0$ , so the absolute value can be removed:

\ln\left(\frac{g-bv(t)}{g}\right)=-bt.

Exponentiate both sides:

\frac{g-bv(t)}{g}=e^{-bt}.

Now isolate $v(t)$ :

g-bv(t)=ge^{-bt},

bv(t)=g-ge^{-bt},

v(t)=\frac{g}{b}\left(1-e^{-bt}\right).

Method 1 for terminal velocity: take the limit as $t\to\infty$ . Since $e^{-bt}\to0$ ,

v_T=\lim_{t\to\infty}v(t)=\frac{g}{b}.

Method 2 for terminal velocity: terminal velocity means acceleration is zero, so plug $dv/dt=0$ into the original differential equation:

0=g-bv_T.

Solving gives

v_T=\frac{g}{b},

which matches the limit from $v(t)$ .

What if instead of linear drag, we use quadratic drag? You may think that it is just as simple as linear drag. However, changing to quadratic drag increases the integration level by a lot, so as an extension and practice to integration, you can try to find $v(t)$ for quadratic drag (hint: you may need to use the hyperbolic tangent function).

Circular Motion

Often, we often describe thing as going in circles (or approximately so), whether it be vertical or horizontal. There are two types of circular motion: uniform and non-uniform circular motion.

Uniform circular motion

In uniform circular motion, speed is constant but velocity changes direction. The acceleration points toward the center at every instant and has a magnitude of

a_c = \frac{v^2}{r}.

Useful variables

Symbol	Meaning
$r$	Radius of the circular path
$v$	Constant speed along the path
$a_c$	Magnitude of the inward, centripetal acceleration
$\theta$	Angular position measured from a chosen reference direction
$\theta_0$	Angular position at $t=0$
$\omega$	Signed angular velocity; positive is counterclockwise
$T$	Period, or time required for one revolution
$f$	Frequency, or revolutions per unit time

Proof (Centripetal acceleration). Start with the parameterization for a standard circle centered at $(a, e)$

x(t)=a+b\cos(ct+d),\qquad y(t)=e+b\sin(ct+d),

where $a,e$ are the center coordinates, $b>0$ is a length, $c\ne0$ has units of inverse time, and $d$ is the starting angle. Since

(x-a)^2+(y-e)^2=b^2,

the path is a circle of radius $b$ . Differentiate both components:

\vec v(t)=-bc\sin(ct+d)\hat i+bc\cos(ct+d)\hat j,

\vec a(t)=-bc^2\cos(ct+d)\hat i-bc^2\sin(ct+d)\hat j.

The magnitude of velocity (speed) is $v=b\lvert c\rvert$ . The acceleration is $-c^2$ times the displacement from the center, so it points inward and has magnitude $bc^2$ .

A complete revolution changes the phase by $2\pi$ . Thus the period (time per cycle) and frequency (cycles per second) are

T=\frac{2\pi}{\lvert c\rvert},\qquad f=\frac{1}{T}=\frac{\lvert c\rvert}{2\pi}.

We call the signed rate of change of angle the angular velocity, $\omega=c$ : positive for counterclockwise motion and negative for clockwise motion. Its magnitude is the angular speed, $\lvert\omega\rvert=2\pi f$ . Writing $b=r$ , $d=\theta_0$ , and the center as $(x_c,y_c)$ gives

x(t)=x_c+r\cos(\omega t+\theta_0),\qquad y(t)=y_c+r\sin(\omega t+\theta_0).

Using $v=r\lvert\omega\rvert$ , the inward acceleration magnitude is

a_c=r\omega^2=\frac{v^2}{r}=\frac{4\pi^2r}{T^2}.

The position relative to the center, velocity, and acceleration are related geometrically: velocity is tangent to the circle, while acceleration is opposite the radius/position vector. “Centripetal” (literally meaning “center-seeking”) is just the name of the inward component of acceleration.

Example. A marker attached to a rigid arm of length $L$ moves in a horizontal circle. The arm stays at angle $\theta$ from the vertical. A camera measures the marker’s constant speed $v$ . Find its period and inward acceleration.

The circle has radius $r=L\sin\theta$ . One revolution covers the circumference of the traveled circle, so

T=\frac{2\pi L\sin\theta}{v},\qquad a_c=\frac{v^2}{L\sin\theta}.

Example. A point moves according to $x=2+3\cos(4t)$ and $y=-1+3\sin(4t)$ , with distances in meters and time in seconds. Find the center, radius, period, speed, and acceleration vector when $t=\pi/8\ \text{s}$ .

The center is $(2,-1)\ \text{m}$ , the radius is $3\ \text{m}$ , and $\omega=4\ \text{rad/s}$ . Therefore

T=\frac{2\pi}{4}=\frac{\pi}{2}\ \text{s},\qquad v=3(4)=12\ \text{m/s},\qquad a_c=3(4)^2=48\ \text{m/s}^2.

At the specified time the phase is $\pi/2$ , so the point is at the top of the circle. Its acceleration points downward:

\vec a=-48\hat j\ \text{m/s}^2.

Nonuniform circular motion

What if the speed of the object is changing while moving in a circle? For example, consider the case of a car decelerating around a turn, a washing machine ending its cycle, or a pendulum swinging about its hinge. For this nonuniform circular motion, acceleration has both radial and tangential components:

a_c = \frac{v^2}{r},

a_t = \frac{dv}{dt}.

The centripetal component $a_c$ points toward the center and changes the direction of velocity. The tangential component $a_t$ lies along the tangent to the circle and changes the magnitude of velocity. If the object is speeding up, $\vec a_t$ points with the velocity; if it is slowing down, $\vec a_t$ points against the velocity.

Using $\hat r$ for the outward radial direction and $\hat\theta$ for the direction of increasing angle, the acceleration vector is

\vec a=-\frac{v^2}{r}\hat r+\frac{dv}{dt}\hat\theta.

The minus sign on the radial term is important: $\hat r$ points outward, while centripetal acceleration points inward. Because the radial and tangential directions are perpendicular, the total acceleration magnitude is

\lvert\vec a\rvert=\sqrt{a_c^2+a_t^2} =\sqrt{\left(\frac{v^2}{r}\right)^2+\left(\frac{dv}{dt}\right)^2}.

If $\phi$ is the angle from the inward radial direction toward the tangential acceleration, then

\tan\phi=\frac{\lvert a_t\rvert}{a_c}.

Notice that even if $a_t$ is constant, $a_c$ generally is not: changing the speed also changes $v^2/r$ .

Example. A particle moves counterclockwise on a circle of radius $R$ , starting at $\theta(0)=0$ with speed $v_0>0$ . Its speed increases according to $dv/dt=\beta v$ , where $\beta>0$ is constant. Given $v_0<\beta R$ , find when its acceleration first makes a $45^\circ$ angle with the inward radius, and find its angular displacement and acceleration vector then.

Separate variables and apply the initial condition:

\frac{dv}{v}=\beta\,dt \quad\Longrightarrow\quad v(t)=v_0e^{\beta t}.

The radial and tangential magnitudes are

a_c=\frac{v_0^2e^{2\beta t}}{R},\qquad a_t=\beta v_0e^{\beta t}.

The angle is $45^\circ$ when these are equal, so the speed must be $v_*=\beta R$ . Therefore

t_*=\frac{1}{\beta}\ln\frac{\beta R}{v_0}.

The angle traveled comes from integrating $\dot\theta=v/R$ :

\theta(t)=\frac{v_0}{\beta R}(e^{\beta t}-1),\qquad \theta_*=1-\frac{v_0}{\beta R}.

At this instant both components have magnitude $\beta^2R$ . Use the outward radial unit vector $\hat r=(\cos\theta,\sin\theta)$ and counterclockwise tangent $\hat\theta=(-\sin\theta,\cos\theta)$ . Thus

\vec a_*=\beta^2R(-\hat r+\hat\theta) =\beta^2R\big[-(\cos\theta_*+\sin\theta_*)\hat i +(\cos\theta_*-\sin\theta_*)\hat j\big],

and $\lvert\vec a_*\rvert=\sqrt2\,\beta^2R$ .

Example. A cyclist rides counterclockwise around a circular track of radius $25\ \text{m}$ . The cyclist’s speed is $v(t)=4.0+0.60t^2$ , where speed is in meters per second and time is in seconds. At $t=3.0\ \text{s}$ , find the tangential acceleration, centripetal acceleration, total acceleration magnitude, and acceleration vector when the cyclist is at the rightmost point of the track.

The tangential acceleration comes from the rate at which the speed changes:

a_t=\frac{dv}{dt}=1.20t.

Therefore, at $t=3.0\ \text{s}$ ,

a_t=1.20(3.0)=3.60\ \text{m/s}^2.

At the same instant, the speed is

v=4.0+0.60(3.0)^2=9.40\ \text{m/s},

so the centripetal acceleration is

a_c=\frac{v^2}{r}=\frac{(9.40)^2}{25}=3.53\ \text{m/s}^2.

The two components are perpendicular, so

\lvert\vec a\rvert=\sqrt{(3.53)^2+(3.60)^2}=5.04\ \text{m/s}^2.

At the rightmost point, inward is left and the counterclockwise tangential direction is up. Thus

\vec a=(-3.53\hat i+3.60\hat j)\ \text{m/s}^2.

The acceleration points

\phi=\tan^{-1}\left(\frac{3.60}{3.53}\right)=45.6^\circ

above the inward radial direction. The nearly equal components do not mean the motion is uniform; the nonzero tangential component shows that the cyclist is speeding up.

Relative velocity (introduction)

When someone is sitting still in a chair, are they moving? In your point of view as an observer in the room with them, they are stationary. Yet, the Earth (and therefore the chair the person is sitting on) constantly rotates around its axis and orbits around the Sun.

To describe one object’s motion as seen from another, choose a reference frame. An inertial frame does not accelerate or rotate relative to another inertial frame; a non-inertial frame does. A frame moving with an object can be either, depending on the object’s motion.

We usually treat the ground-based lab frame as approximately inertial, ignoring Earth’s rotation and orbital acceleration. Watching an F1 race from the stands is an example of this lab-frame viewpoint.

A frame attached to a racecar is non-inertial while the car speeds up, slows down, or turns. In that frame, the driver is at rest and the spectators move relative to the car. Both descriptions are valid; they use different reference frames.

Velocities are vectors, so relative velocities add by vector addition. Define the velocity of object $A$ relative to object $B$ as $\vec{v}_{A/B}$ . To find the velocity relative to a third object C, express all vectors using the same axis directions and write

\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}.

If you need a reminder of vector addition, check out AP Precalculus. Since we can model the velocities as vectors, $\vec{v}_{A/B} = -\vec{v}_{B/A}$ , meaning that two objects that are moving have opposite relative velocities with respect to each other.

What if instead of looking at velocity, we look at acceleration? As long as the reference frames use axes that remain parallel and do not rotate relative to one another,

\vec a_{A/C}=\vec a_{A/B}+\vec a_{B/C}.

Equivalently, the acceleration of $A$ as measured by $B$ is

\vec a_{A/B}=\vec a_{A/C}-\vec a_{B/C}.

An object released from an accelerating frame keeps the velocity it had at the instant of release, but it does not automatically keep the frame’s acceleration. For example, a donut dropped inside an eastward-accelerating train initially has the train’s eastward velocity. After release, however, no horizontal force acts on the donut. A ground observer therefore measures zero horizontal acceleration, while an observer on the train sees the donut accelerate backward relative to the train.

Example. A train moves east at speed $u$ and accelerates east at a constant rate $A$ . At $t=0$ , a passenger releases a donut from rest relative to the train at height $H$ above the floor. Ignore air resistance. Describe the donut’s motion in the ground frame and the train frame, and find how far behind the release point it lands as measured in the train.

Take east as $+x$ and up as $+y$ . Let $D$ denote the donut, $T$ the train, and $G$ the ground. At release, the donut has the train’s instantaneous horizontal velocity:

\vec v_{D/G}(0)=u\hat i.

Once released, the donut is in free fall. In the ground frame its acceleration is

\vec a_{D/G}=-g\hat j,

so its coordinates relative to the release point are

x_{D/G}=ut, \qquad y_{D/G}=H-\frac{1}{2}gt^2.

The ground observer sees a projectile: the donut continues east while falling in a parabola. During the same time, the point on the train directly below the release point moves according to

x_{T/G}=ut+\frac{1}{2}At^2.

Subtracting the train’s position from the donut’s position gives the horizontal motion seen by the passenger:

x_{D/T}=x_{D/G}-x_{T/G}=-\frac{1}{2}At^2.

Thus the donut appears to accelerate west in the train frame, with

\vec a_{D/T}=\vec a_{D/G}-\vec a_{T/G}=-A\hat i-g\hat j.

The donut reaches the floor when $y_{D/G}=0$ :

t_f=\sqrt{\frac{2H}{g}}.

At that time,

x_{D/T}(t_f) =-\frac{1}{2}A\left(\frac{2H}{g}\right) =-\frac{AH}{g}.

Therefore, the donut lands a distance $AH/g$ behind the point on the train directly below where it was released. The two observers agree on the physical landing event, but they describe the path and acceleration differently because the train frame is accelerating.

Example. A boat heads straight across a river, pointing its bow perpendicular to the banks with a speed of $4\ \text{m/s}$ relative to the water. The river flows at $3\ \text{m/s}$ parallel to the banks, and the river is $80\ \text{m}$ wide. Find the boat’s velocity relative to the ground, how long the crossing takes, and how far downstream it lands.

Let the boat’s velocity relative to the water be $\vec{v}_{B/W}$ (across the river) and the water’s velocity relative to the ground be $\vec{v}_{W/G}$ (downstream). The composition rule gives

\vec{v}_{B/G} = \vec{v}_{B/W} + \vec{v}_{W/G}.

These two pieces are perpendicular (check it yourself), so the ground speed is

v_{B/G} = \sqrt{4^2 + 3^2} = 5\ \text{m/s},

at an angle $\arctan(3/4) \approx 37^\circ$ downstream of straight across.

The crossing time depends only on the across-river component, because the downstream flow does nothing to close the $80\ \text{m}$ gap:

t = \frac{80}{4} = 20\ \text{s}.

During that time the current carries the boat downstream by

d = (3)(20) = 60\ \text{m}.

The key idea is that the perpendicular components are independent, just as in projectile motion: the downstream drift does not change how long the crossing takes.

You can find more about relative motion on the USAPhO section on mechanics.

Kinematics in two dimensions and projectile motion

In two-dimensional kinematics, a vector can be split into $x$ - and $y$ -components, and each component follows its own kinematic equation. Projectile motion is the motion of an object after it has been launched and is moving under gravity alone. If air resistance is negligible, horizontal acceleration is zero and vertical acceleration is $g$ downward. The sign of the vertical acceleration depends on which direction you choose as positive $y$ .

With initial speed $v_0$ at launch angle $\theta$ above the horizontal,

v_{0x} = v_0 \cos\theta, v_{0y} = v_0 \sin\theta.

The motion can thus be broken into components, each with its own kinematic equations. Taking up as positive $y$ gives

x = x_0 + v_{0x} t, \qquad y = y_0 + v_{0y} t - \frac{1}{2} g t^2,

v_x = v_{0x}, \qquad v_y = v_{0y} - gt.

Remember: ALWAYS make sure you know which direction you define as $+y$ ! The trajectory in the vertical plane is a parabola until the object hits something.

If launch and landing occur at the same height, you can shortcut formulas. The shortcuts for range, maximum height, and total flight time on level ground are:

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

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

T = \frac{2v_0 \sin(\theta)}{g}.

If launch and landing heights differ, solve the quadratic in $t$ from the $y$ equation rather than memorizing these shortcuts.

Proof (Range, Max Height, Flight Time). Take up as $+y$ , launch from the origin, and write the component equations:

y = v_0\sin\theta\, t - \tfrac{1}{2}g t^2, \qquad x = v_0\cos\theta\, t,

v_y = v_0\sin\theta - gt.

Flight time. On level ground the projectile lands when $y = 0$ again:

0 = v_0\sin\theta\, T - \tfrac{1}{2}g T^2 = T\left(v_0\sin\theta - \tfrac{1}{2}g T\right).

The root $T = 0$ is the launch instant; the landing is the other root:

T = \frac{2 v_0 \sin\theta}{g}.

Maximum height. At the top of the arc the vertical velocity is zero, $v_y = 0$ , which happens at $t = v_0\sin\theta / g$ , exactly half of $T$ (the parabola is symmetric). Substitute into $y$ :

h = v_0\sin\theta\left(\frac{v_0\sin\theta}{g}\right) - \tfrac{1}{2}g\left(\frac{v_0\sin\theta}{g}\right)^2 = \frac{v_0^2\sin^2\theta}{2g}.

Range. The horizontal distance covered in the full flight time is

R = v_0\cos\theta\cdot T = v_0\cos\theta\cdot\frac{2v_0\sin\theta}{g} = \frac{2v_0^2\sin\theta\cos\theta}{g}.

Using the identity $2\sin\theta\cos\theta = \sin(2\theta)$ ,

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

Because $\sin(2\theta)$ is largest at $2\theta = 90^\circ$ , range on level ground is maximized at $\theta = 45^\circ$ , and complementary angles such as $30^\circ$ and $60^\circ$ give the same range.

Example. A ball is launched from level ground at $v_0 = 20\ \text{m/s}$ and $\theta = 30^\circ$ . Take $g = 9.8\ \text{m/s}^2$ . Find the time of flight, range, and maximum height.

With $\sin 30^\circ = 0.5$ , $\cos 30^\circ \approx 0.866$ , and $\sin 60^\circ \approx 0.866$ :

T = \frac{2(20)(0.5)}{9.8} \approx 2.0\ \text{s},

R = \frac{(20)^2(0.866)}{9.8} \approx 35.4\ \text{m},

h = \frac{(20)^2(0.5)^2}{2(9.8)} \approx 5.1\ \text{m}.

As a sanity check, a $60^\circ$ launch at the same speed would give the same range $R$ but a much greater height, since $\sin^2 60^\circ = 0.75$ is three times $\sin^2 30^\circ$ .

Example. A projectile is launched from the top of a cliff $45\ \text{m}$ tall at $v_0 = 30\ \text{m/s}$ and $\theta = 37^\circ$ above the horizontal. Take $g = 10\ \text{m/s}^2$ , $\sin 37^\circ = 0.6$ , $\cos 37^\circ = 0.8$ . How long is it in the air, and how far from the base of the cliff does it land?

Put the origin at the launch point with up as $+y$ . Then the components are

v_{0x} = 30(0.8) = 24\ \text{m/s}, \qquad v_{0y} = 30(0.6) = 18\ \text{m/s}.

The ground is $45\ \text{m}$ below launch, so landing is at $y = -45\ \text{m}$ :

-45 = 18t - \tfrac{1}{2}(10)t^2 = 18t - 5t^2.

Rearranging into standard form,

5t^2 - 18t - 45 = 0.

Apply the quadratic formula:

t = \frac{18 \pm \sqrt{(-18)^2 - 4(5)(-45)}}{2(5)} = \frac{18 \pm \sqrt{324 + 900}}{10} = \frac{18 \pm 35}{10}.

The physical (positive) root is

t = \frac{18 + 35}{10} = 5.3\ \text{s}.

The negative root is discarded because it corresponds to a time before launch. The horizontal range from the base of the cliff is

x = v_{0x}\, t = 24(5.3) \approx 127\ \text{m}.

Note that with unequal launch and landing heights the trajectory is no longer symmetric, which is why the level-ground shortcuts cannot be used here.

Inclined plane kinematics

Suppose we want to launch along a surface that is angle $\theta$ above the horizontal. For a projectile above a ramp, rotate the axes: let $x$ point along the ramp to the right and $y$ point perpendicular to it, away from the surface. Let $\theta$ be the signed ramp angle relative to horizontal: positive for an uphill ramp and negative for a downhill slope, with $-90^\circ<\theta<90^\circ$ . Launch from the origin at speed $v_0$ and angle $\phi$ above the ramp, so the launch angle above horizontal is $\theta+\phi$ .

Gravity is the only acceleration during flight. Its rotated components and the velocity components are

a_x=-g\sin\theta,\qquad a_y=-g\cos\theta,

v_x=v_0\cos\phi-gt\sin\theta,\qquad v_y=v_0\sin\phi-gt\cos\theta.

Integrating with initial position at the origin gives

x=v_0t\cos\phi-\frac12gt^2\sin\theta,\qquad y=v_0t\sin\phi-\frac12gt^2\cos\theta.

The ramp itself is $y=0$ in these coordinates. The projectile leaves the surface, so its perpendicular coordinate is positive during flight; it is not constrained to slide along the ramp. For a downhill slope, $\theta<0$ and $a_x>0$ , so gravity increases the down-slope velocity component. You can try to prove the formulas by yourself (hint: Use similar triangles and Pythagorean Theorem).

Example. A projectile is launched at speed $v_0$ above an uphill ramp of angle $\theta$ , where $0<\theta<90^\circ$ . Its launch direction makes angle $\phi$ above the ramp, with $0<\phi<90^\circ-\theta$ . Find its landing distance $d$ along the ramp and the launch angle that maximizes this distance.

Landing occurs when the perpendicular displacement returns to zero:

0=t\left(v_0\sin\phi-\frac12gt\cos\theta\right).

Discarding the launch instant $t=0$ gives

t_f=\frac{2v_0\sin\phi}{g\cos\theta}.

Substitute into the along-ramp position:

d=\frac{2v_0^2\sin\phi\cos\phi}{g\cos\theta} -\frac{2v_0^2\sin^2\phi\sin\theta}{g\cos^2\theta} =\frac{2v_0^2\sin\phi\cos(\theta+\phi)}{g\cos^2\theta}.

Using $2\sin\phi\cos(\theta+\phi)=\sin(2\phi+\theta)-\sin\theta$ ,

d=\frac{v_0^2}{g\cos^2\theta} \left[\sin(2\phi+\theta)-\sin\theta\right].

Only the sine term varies with launch angle. It is largest when $2\phi+\theta=90^\circ$ , so

\phi_{\mathrm{opt}}=45^\circ-\frac{\theta}{2},\qquad d_{\max}=\frac{v_0^2}{g(1+\sin\theta)}.

The optimum angle above horizontal is therefore $45^\circ+\theta/2$ . For a downhill slope of angle $\beta>0$ , substitute $\theta=-\beta$ : the optimum angle above the slope is $45^\circ+\beta/2$ .

If instead a reachable distance $d$ is specified, there are generally two launch angles. Set

K=\sin\theta+\frac{gd\cos^2\theta}{v_0^2}.

For $0<d<d_{\max}$ , the larger angle above the ramp that reaches that distance is

\phi_{\mathrm{high}}=\frac{180^\circ-\arcsin K-\theta}{2}.

At $d=d_{\max}$ the two angles coincide. A requested distance greater than $d_{\max}$ is unreachable at that launch speed.

Calculus method. Hold $v_0$ and $\theta$ fixed and differentiate the range with respect to $\phi$ , measuring angles in radians:

\frac{dd}{d\phi}=\frac{2v_0^2}{g\cos^2\theta}\cos(2\phi+\theta).

Setting this equal to zero gives $2\phi+\theta=\pi/2$ in the allowed launch-angle interval. Thus

\phi_{\mathrm{opt}}=\frac{\pi}{4}-\frac{\theta}{2}.

To confirm that this critical point maximizes the range, differentiate again:

\frac{d^2d}{d\phi^2} =-\frac{4v_0^2}{g\cos^2\theta}\sin(2\phi+\theta).

At the critical point, the sine is $1$ , so the second derivative is negative. The range also approaches zero at both endpoints $\phi=0$ and $\phi=\pi/2-\theta$ of the uphill launch interval, making this the absolute maximum.

Key equations

Idea	Equation
Average velocity / acceleration	$\bar{v} = \dfrac{\Delta x}{\Delta t},\quad \bar{a} = \dfrac{\Delta v}{\Delta t}$
Instantaneous velocity	$v = \dfrac{dx}{dt}$
Instantaneous acceleration	$a = \dfrac{dv}{dt} = \dfrac{d^2x}{dt^2}$
Velocity / position by integration	$\Delta v = \displaystyle\int a\,dt,\quad \Delta x = \displaystyle\int v\,dt$
Acceleration depending on position	$a = v\dfrac{dv}{dx}$
Constant accel: missing $\Delta x$	$v_f = v_0 + at$
Constant accel: missing $v_f$	$\Delta x = v_0 t + \tfrac{1}{2}at^2$
Constant accel: missing $v_0$	$\Delta x = v_f t - \tfrac{1}{2}at^2$
Constant accel: missing $t$	$v_f^2 = v_0^2 + 2a\Delta x$
Constant accel: missing $a$	$\Delta x = \dfrac{v_0 + v_f}{2}\,t$
Projectile launch components	$v_{0x} = v_0\cos\theta,\quad v_{0y} = v_0\sin\theta$
Projectile motion (up is $+y$ )	$x = v_{0x}t,\quad y = v_{0y}t - \tfrac{1}{2}gt^2$
Projectile velocity	$v_x = v_{0x},\quad v_y = v_{0y} - gt$
Range (level ground)	$R = \dfrac{v_0^2\sin(2\theta)}{g}$
Max height (level ground)	$h = \dfrac{v_0^2\sin^2\theta}{2g}$
Time of flight (level ground)	$T = \dfrac{2v_0\sin\theta}{g}$
Relative velocity	$\vec{v}_{A/C} = \vec{v}_{A/B} + \vec{v}_{B/C}$
Relative acceleration (parallel, nonrotating axes)	$\vec a_{A/C}=\vec a_{A/B}+\vec a_{B/C}$
Circular speed, period, frequency	$v=r\lvert\omega\rvert=2\pi r/T,\quad f=1/T$
Circular acceleration components	$a_c=v^2/r=r\omega^2,\quad a_t=dv/dt$
Total circular acceleration	$\lvert\vec a\rvert=\sqrt{a_c^2+a_t^2}$
Linear drag acceleration (down positive, released from rest)	$dv/dt=g-bv,\quad v(t)=\frac gb(1-e^{-bt}),\quad v_T=g/b$
Ramp axes: signed slope angle $\theta$ , launch angle $\phi$ above ramp	$a_x=-g\sin\theta,\quad a_y=-g\cos\theta$
Ramp velocity components	$v_x=v_0\cos\phi-gt\sin\theta,\quad v_y=v_0\sin\phi-gt\cos\theta$
Ramp position from launch	$x=v_0t\cos\phi-\frac12gt^2\sin\theta,\quad y=v_0t\sin\phi-\frac12gt^2\cos\theta$
Return to ramp	$t_f=\frac{2v_0\sin\phi}{g\cos\theta},\quad d=\frac{v_0^2[\sin(2\phi+\theta)-\sin\theta]}{g\cos^2\theta}$
Maximum ramp range (fixed launch speed)	$\phi_{\mathrm{opt}}=\pi/4-\theta/2,\quad d_{\max}=\frac{v_0^2}{g(1+\sin\theta)}$

Practice

Multiple Choice

A projectile is launched from level ground. At the top of its path, its speed is half its launch speed. What was the launch angle?

(A) $30^\circ$

(B) $45^\circ$

(D) $75^\circ$

Two projectiles are launched from the same point with the same speed at complementary angles $\theta$ and $90^\circ-\theta$ , where $0<\theta<45^\circ$ . On level ground, the projectile launched at the larger angle has

(A) the same range and a longer flight time

(B) the same range and a shorter flight time

(D) a shorter range and a shorter flight time

A particle has $x(t)=At^3-Bt$ with $A,B>0$ . For $t>0$ , at the instant when the particle’s velocity is zero, its acceleration is

(A) zero

(B) $-2\sqrt{3AB}$

(D) $6A\sqrt{B/(3A)}$

A particle moves in the plane with $x=bt$ and $y=ct^2-dt^3$ , where $b,c,d>0$ . At the nonzero instant when $v_y=0$ , the acceleration vector points

(A) purely horizontal

(B) upward

(D) tangent to the trajectory

A runner moves so that her speed depends on position according to $v=v_0+kx$ , where $v_0,k>0$ . Her acceleration as a function of position is

(A) $k$

(B) $k(v_0+kx)$

(D) $v_0+kx$

A particle has $v(t)=v_0-\beta t^2$ with $v_0,\beta>0$ . Which expression gives the distance traveled from $t=0$ until the particle first stops?

(A) $\int_0^{\sqrt{v_0/\beta}}(v_0-\beta t^2)\,dt$

(B) $\int_0^{v_0/\beta}(v_0-\beta t^2)\,dt$

(D) $v_0\sqrt{v_0/\beta}$

A particle moves along the $x$ -axis with velocity $v(t)=v_0-\alpha t^2$ , where $v_0,\alpha>0$ . At what time is the particle’s displacement from its starting point greatest?

(A) $t=\sqrt{v_0/\alpha}$

(B) $t=v_0/\alpha$

(D) $t=2v_0/\alpha$

A projectile is launched from level ground and lands back at the same height a fixed horizontal distance $R$ away. The launch speed is increased while $R$ is kept the same. Compared with the original two possible launch angles, the new two possible launch angles

(A) move closer to $45^\circ$

(B) move farther from $45^\circ$

(D) both decrease

A boat always points directly across a river of width $W$ with speed $v_b$ relative to the water. The current is parallel to the banks and has speed $u(y)=u_0y/W$ , where $y$ is distance across the river. Compared with a river whose current is everywhere $u_0/2$ , the boat’s downstream drift is

(A) smaller

(B) the same

(D) impossible to compare without $v_b$

A particle moves in one dimension with acceleration $a=-kv^2$ when $v>0$ , where $k>0$ . Which statement must be true while the particle is moving in the positive direction?

(A) The velocity-time graph is a straight line.

(B) The velocity decreases, but the magnitude of the slope decreases as the particle slows.

(D) Equal decreases in speed take equal amounts of time.

A skier launches from a point on a long slope descending at $30^\circ$ below horizontal. The launch speed is $v_0$ , and the launch angle above the slope is chosen to maximize the distance along the slope before landing. Ignore air resistance. Which pair gives the maximum distance $d$ and the speed immediately before landing?

(A) $d=v_0^2/g,\quad v=v_0$

(B) $d=2v_0^2/g,\quad v=\sqrt{3}v_0$

(D) $d=3v_0^2/g,\quad v=\sqrt{3}v_0$

A small cart moves to the right with initial speed $v_0$ through a medium that produces resistive acceleration

a=-kv\left(1+\frac{v}{V}\right)

while $v>0$ , where $k>0$ and $V>0$ are constants. Which expression gives the cart’s velocity?

(A) $v(t)=\dfrac{Vv_0e^{-kt}}{V+v_0\left(1-e^{-kt}\right)}$

(B) $v(t)=\dfrac{Vv_0}{V+kv_0t}$

(D) $v(t)=V\left(e^{-kt}-1\right)+v_0$

FRQ

A bead moves along a straight track with acceleration $a(x)=\alpha x-\beta$ , where $\alpha$ and $\beta$ are positive constants. At $x=0$ , the bead has speed $v_0$ in the positive direction.

$(A)$ Derive an expression for $v^2$ as a function of $x$ .

$(B)$ Find the condition on $v_0$ for the bead to reach $x=\beta/\alpha$ .

$(C)$ If the bead turns around before reaching $x=\beta/\alpha$ , determine the turning point.

$(D)$ Explain how the result changes if the bead initially moves in the negative direction.

A projectile is launched from a cliff of height $H$ with initial speed $v_0$ at angle $\theta$ above horizontal. A horizontal wind causes constant acceleration $a_w$ in the same direction as the projectile’s horizontal velocity.

$(A)$ Derive expressions for $x(t)$ and $y(t)$ .

$(B)$ Find an equation for the time when the projectile reaches the ground.

$(C)$ Derive the horizontal distance from the base of the cliff where the projectile lands.

$(D)$ Determine whether increasing $a_w$ changes the time of flight, and justify your answer.

A particle moves along the $x$ -axis. From $t=0$ to $t=T$ , its velocity is $v(t)=v_0(1-t/T)^2$ . From $t=T$ to $t=2T$ , its acceleration is constant and chosen so the particle returns to its starting position at $t=2T$ .

$(A)$ Find the displacement during the first interval.

$(B)$ Determine the velocity at $t=T$ .

$(C)$ Find the constant acceleration during the second interval.

$(D)$ Sketch the velocity-time graph, labeling intercepts and areas with signs.