AP Physics C Mechanics: Lecture Notes

These are the notes of my AP Physics C: Mechanics lectures. AP Physics C uses calculus from the first day: velocity is the derivative of position, and displacement is the integral of velocity. So the course starts with kinematics and the little calculus it needs, and each new tool is introduced when the physics asks for it. The examples are the ones we worked through on the board.

Unit	Topic	Status
1	Kinematics, vectors and projectile motion	in progress
2	Force and translational dynamics	coming later
3	Work, energy and power	coming later
4	Linear momentum	coming later
5	Torque and rotational dynamics	coming later
6	Energy and momentum of rotating systems	coming later
7	Oscillations	coming later

1. Kinematics in One Dimension

We study motion along a straight line. We choose an origin $0$ and a positive direction on the line; positions to the right of the origin are positive, positions to the left are negative.

Position, Velocity and Acceleration

Definition. For a body moving on a line,

$x (t)$ , the position, is a function of time $t$ ;
$v (t)$ , the velocity, is a function of $t$ ;
$a (t)$ , the acceleration, is a function of $t$ .

In the first part of the course the acceleration is constant: $a (t) = a$ for one fixed number $a$ , such as $3$ , $5$ , $0$ , $- 2$ or $- π$ .

Definition. The velocity is the derivative of the position: $v (t) = \frac{d x}{d t} .$

Even on a line, velocity is a vector. In one dimension its direction is only a sign: $+$ means the body moves in the positive direction, $-$ means it moves in the negative direction.

Average and Instantaneous Velocity

Definition. The average velocity between the times $t_{1}$ and $t_{2}$ is $v_{avg} = \frac{Δ x}{Δ t} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}},$ the slope of the secant line through the points $(t_{1}, x_{1})$ and $(t_{2}, x_{2})$ of the position graph.

Example. Let $x (t) = t^{2}$ . What is the average velocity between $t = 3$ and $t = 4$ ?

$v_{avg} = \frac{4^{2} - 3^{2}}{4 - 3} = \frac{7}{1} = 7.$

Now keep $t_{1} = 3$ and move $t_{2}$ closer to $3$ :

$t_{1}$	$t_{2}$	$v_{avg} = Δ x / Δ t$
$3$	$4$	$7$
$3$	$3.5$	$6.5$
$3$	$3.1$	$6.1$
$3$	$3.01$	$6.01$
$⋮$	$\to 3$	$\to 6$

With $Δ x = t_{2}^{2} - 9$ , $lim_{t_{2} \to 3} \frac{t_{2}^{2} - 9}{t_{2} - 3} = lim_{t_{2} \to 3} \frac{(t_{2} - 3) (t_{2} + 3)}{t_{2} - 3} = lim_{t_{2} \to 3} (t_{2} + 3) = 6.$

The graph of x = t squared with the secant line from t = 3 to t = 4, of slope 7, and the tangent line at t = 3, of slope 6 — As the second point slides down to $t = 3$ , the secant lines turn into the tangent line at $t = 3$ .

This is the link between average and instantaneous velocity: the derivative.

Definition. The instantaneous velocity is the limit of the average velocity as $Δ t \to 0$ : $v (t) = \frac{d x}{d t} = lim_{Δ t \to 0} \frac{Δ x}{Δ t} .$ It is the slope of the tangent line to the position graph.

The Power Rule

Theorem (Power rule). If $x (t) = t^{n}$ , then $\frac{d x}{d t} = x^{'} (t) = n t^{n - 1} .$

Example. For the motion above, $x (t) = t^{2}$ , so $\frac{d x}{d t} = 2 t$ and $v (t) = 2 t$ . At $t = 3$ , $v (3) = {\frac{d x}{d t} |}_{t = 3} = 2 \cdot 3 = 6,$ the same value as the limit of the table.

Acceleration

Definition. The average acceleration is $a_{avg} = \frac{Δ v}{Δ t}$ , and the instantaneous acceleration is the derivative of the velocity: $a (t) = \frac{d v}{d t} .$

For $x (t) = t^{2}$ we get $v (t) = 2 t$ and $a (t) = 2$ .

Derivative and Integral: Reading the Graphs

Going down from $x (t)$ to $v (t)$ to $a (t)$ we take derivatives. Going up we take integrals:

the area under the $v$ – $t$ graph between $t_{1}$ and $t_{2}$ is the change in position, $x_{2} - x_{1} = \int_{t_{1}}^{t_{2}} v (t) d t$ ;
the area under the $a$ – $t$ graph between $t_{1}$ and $t_{2}$ is the change in velocity, $v_{2} - v_{1} = \int_{t_{1}}^{t_{2}} a (t) d t$ .

Area below the $t$ -axis counts as negative.

General Formulas

Summary. $v (t) = lim_{Δ t \to 0} \frac{Δ x}{Δ t} = \frac{d x}{d t} ⟺ Δ x = \int d x = \int_{t_{1}}^{t_{2}} v (t) d t$

$a (t) = lim_{Δ t \to 0} \frac{Δ v}{Δ t} = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}} ⟺ Δ v = \int d v = \int_{t_{1}}^{t_{2}} a (t) d t$

Speed is the absolute value of the velocity: $| v (t) |$ .
Average velocity: $\bar{v} = \frac{1}{t_{2} - t_{1}} \int_{t_{1}}^{t_{2}} v (t) d t = \frac{x (t_{2}) - x (t_{1})}{t_{2} - t_{1}} = \frac{Δ x}{Δ t}$ .
Average speed: $\frac{total distance}{Δ t}$ .
Average acceleration: $\bar{a} = \frac{1}{t_{2} - t_{1}} \int_{t_{1}}^{t_{2}} a (t) d t = \frac{v (t_{2}) - v (t_{1})}{t_{2} - t_{1}} = \frac{Δ v}{Δ t}$ .

$x (t) \underset{integrate}{\overset{differentiate}{⇄}} v (t) \underset{integrate}{\overset{differentiate}{⇄}} a (t)$

Remark. Displacement and distance are different. A body that goes from $x = 0$ to $x = - 10$ has displacement $Δ x = - 10 - 0 = - 10$ , but the distance it travelled is $10$ .

Derivative Examples

Function	Derivative
$t^{2}$	$2 t$
$t$	$1$
$3 t^{2}$	$6 t$
$- 2 t$	$- 2$
a constant: $1, 2, 3, 5, 0, \dots$	$0$
$\frac{1}{t} = t^{- 1}$	$(- 1) t^{- 2} = - \frac{1}{t^{2}}$
$- \frac{1}{t^{3}} = - t^{- 3}$	$- (- 3) t^{- 4} = \frac{3}{t^{4}}$
$\sqrt{t} = t^{1 / 2}$	$\frac{1}{2} t^{- 1 / 2} = \frac{1}{2 \sqrt{t}}$
$2 t^{2} + 3 t$	$4 t + 3$
$\sin t + t^{2}$	$\cos t + 2 t$

Remark. A derivative of a sum is the sum of the derivatives, but a derivative of a product is not the product of the derivatives: $(\sin t \cdot t^{2})^{'} \neq \cos t \cdot 2 t$ . For products we use the product rule, "derivative of the first times the second plus the first times the derivative of the second": $(\sin t \cdot t^{2})^{'} = \cos t \cdot t^{2} + \sin t \cdot 2 t .$

Example. A body moves with $x (t) = 2 t - 3 t^{2}$ . Find its position, velocity, speed and acceleration at $t = 1$ .

$v (t) = \frac{d x}{d t} = 2 - 6 t, a (t) = \frac{d v}{d t} = - 6.$

So $x (1) = 2 - 3 = - 1$ , $v (1) = - 4$ , $a (1) = - 6$ , and the speed is $| v (1) | = | - 4 | = 4$ .

Equations of Motion with Constant Acceleration

Assume that the moving body is at $x_{0}$ at $t = 0$ and has the velocity $v_{0}$ at $t = 0$ .

Example. Let $a (t) = 3$ , $x (0) = 3$ and $v (0) = 1$ . A function whose derivative is $3$ is $3 t + k$ , where $k$ can be any number: $v (t) = 3 t + k, v (0) = 1 \Rightarrow 3 \cdot 0 + k = 1 \Rightarrow k = 1, v (t) = 3 t + 1.$ In the same way $x (t) = \frac{3}{2} t^{2} + t + c, x (0) = 3 \Rightarrow c = 3, x (t) = \frac{3}{2} t^{2} + t + 3.$

Summary (Constant acceleration). In general, with $x (0) = x_{0}$ , $v (0) = v_{0}$ and $a (t) = a$ : $v (t) = v_{0} + a t, x (t) = x_{0} + v_{0} t + \frac{1}{2} a t^{2} .$ Two more formulas follow from these: $v^{2} = v_{0}^{2} + 2 a (x - x_{0}), \bar{v} = \frac{v + v_{0}}{2} .$ In terms of changes: $Δ v = a t$ , $Δ x = v_{0} t + \frac{1}{2} a t^{2}$ and $v_{f}^{2} - v_{i}^{2} = 2 a Δ x$ .

Example. A body starts from rest at the origin with acceleration $a = 3 {m/s}^{2}$ . Find its velocity and position after $3$ s.

$x_{0} = 0$ , $v_{0} = 0$ , $a = 3$ , so $v (t) = 3 t$ and $x (t) = \frac{1}{2} \cdot 3 t^{2}$ . Then $v (3) = 9 m/s$ and $x (3) = 13.5 m$ .

Example. A body has $v (0) = 3$ , $x (0) = - 7$ and $a = 1$ . What is its average velocity between $t = 3$ and $t = 4$ ?

$x (4) = - 7 + 3 \cdot 4 + \frac{1}{2} \cdot 1 \cdot 4^{2} = 13, x (3) = - 7 + 3 \cdot 3 + \frac{1}{2} \cdot 1 \cdot 3^{2} = 6.5,$ $\bar{v} = \frac{Δ x}{Δ t} = \frac{x (4) - x (3)}{4 - 3} = \frac{13 - 6.5}{1} = 6.5 .$

Graphs of $x$ , $v$ and $a$

Example. For $x (t) = t^{2}$ we have $v (t) = 2 t$ and $a (t) = 2$ . The position graph is a parabola, the velocity graph is a line through the origin, and the acceleration graph is a horizontal line. Between $t = 0$ and $t = 1$ , the area under the $v$ -graph is the triangle $\frac{1}{2} \cdot 1 \cdot 2 = 1 = x (1) - x (0)$ , and the area under the $a$ -graph is the rectangle $1 \cdot 2 = 2 = v (1) - v (0)$ .

Example. The acceleration of a body is $2$ on $[0, 1]$ , $1$ on $[1, 2]$ and $- 3$ on $[2, 3]$ . If $v_{0} = 0$ and $x (0) = 0$ , find $v (3)$ and $x (3)$ .

The areas under the $a$ -graph are $2$ , $1$ and $- 3$ , so the velocity grows to $v (1) = 2$ , then to $v (2) = 3$ , and falls back to $v (3) = 0 + 2 + 1 - 3 = 0.$ The velocity graph is made of straight pieces, and $x (3)$ is the total area under it: $x (3) = \underset{1}{\underset{⏟}{\frac{1}{2} \cdot 1 \cdot 2}} + \underset{2.5}{\underset{⏟}{\frac{2 + 3}{2} \cdot 1}} + \underset{1.5}{\underset{⏟}{\frac{1}{2} \cdot 1 \cdot 3}} = 5.$

Left: acceleration 2, 1 and minus 3 on three one-second intervals. Right: the velocity graph rising from 0 to 2 to 3 and falling back to 0, with areas 1, 2.5 and 1.5 under it — Areas under $a (t)$ build $v (t)$ ; areas under $v (t)$ build $x (t)$ .

Practice Problems

Exercise (1). A body moving in the positive $x$ direction passes the origin at $t = 0$ . From $t = 0$ to $t = 1$ s it moves with the constant speed $24$ m/s. At $t = 1$ s it is given a constant acceleration of $6 {m/s}^{2}$ in the negative $x$ direction. Where is it at $t = 11$ s?

Solution. For $a = 0$ and $x_{0} = 0$ the formula $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ gives $x = v t$ , so from $0$ to $1$ s the body travels $24$ m. From $1$ s to $11$ s it starts at $24$ m with $v_{0} = 24$ m/s and $a = - 6 {m/s}^{2}$ for $10$ s: $x = 24 + 24 \cdot 10 + \frac{1}{2} (- 6) (10)^{2} = 24 + 240 - 300 = - 36 m .$

Exercise (2). A position–time graph of a body moving under a constant force shows $x = 2$ m at $t = 1$ s and $x = 8$ m at $t = 2$ s. The body does not turn back in between. What is its average speed between $1$ s and $2$ s?

Solution. $\bar{v} = \frac{Δ x}{Δ t} = \frac{8 - 2}{1} = 6 m/s .$

Exercise (3). A body is uniformly accelerated from rest. What do the graphs of distance against time and speed against time look like?

Solution. With $v_{0} = 0$ we have $x = \frac{1}{2} a t^{2}$ and $v = a t$ . The distance graph is a parabola that opens upward (it gets steeper and steeper), and the speed graph is a straight line through the origin.

Exercise (4). The velocity of a body moving on a line rises steadily from $0$ , stays constant for a while, then falls steadily back to $0$ . Describe its position graph.

Solution. The slope of the position graph is the velocity. While the velocity grows, the position graph gets steeper (it bends upward); while the velocity is constant, it is a straight line; while the velocity decreases to $0$ , it gets flatter (it bends downward) and ends horizontal. The position keeps increasing the whole time, because the velocity is never negative.

2. A Little Calculus: Derivatives and Integrals

Taking the derivative goes from the function to its derivative; integration goes backwards, from the derivative to the function.

Derivatives of Basic Functions

Function	Derivative
a constant $k$ : $1, π, e, 0, \dots$	$0$
$t$	$1$
$t^{n}$ (power rule)	$n t^{n - 1}$
$\sin t$	$\cos t$
$\cos t$	$- \sin t$
$\tan t$	$\sec^{2} t$
$e^{t}$	$e^{t}$ (its own derivative)
$\ln t$	$\frac{1}{t} = t^{- 1}$

Examples of the Power Rule

$x (t) = t^{3}$ : $x^{'} (t) = v (t) = 3 t^{2}$ .
$v (t) = t^{1 / 2}$ : $a = \frac{d v}{d t} = \frac{1}{2} t^{- 1 / 2} = \frac{1}{2 \sqrt{t}}$ .
$x (t) = t^{- 3}$ : $v (t) = \frac{d x}{d t} = (- 3) t^{- 4} = - \frac{3}{t^{4}}$ , $a (t) = \frac{d v}{d t} = (- 4) (- 3) t^{- 5} = \frac{12}{t^{5}}$ .

Remark. Read the power rule backwards. To get the derivative $t^{2}$ we need $\frac{t^{3}}{3}$ ; to get $t$ we need $\frac{t^{2}}{2}$ ; to get $1$ we need $t$ ; to get $t^{- 2}$ we need $- t^{- 1}$ .

Function	Derivative	Power of the derivative
$t^{3} / 3$	$t^{2}$	$2$
$t^{2} / 2$	$t$	$1$
$t$	$1$	$0$
$t^{0} = 1$	$0$	(none)
$- t^{- 1}$	$t^{- 2}$	$- 2$

No power $t^{n}$ has the derivative $t^{- 1}$ : the power $- 1$ is missing from the last column. The function whose derivative is $t^{- 1} = \frac{1}{t}$ is $\ln t$ .

Basic Rules of Differentiation

Constant multiple: $\frac{d}{d t} (3 f (t)) = 3 \frac{d f}{d t}$ . For example ${(\frac{1}{3} t^{4})}^{'} = \frac{4}{3} t^{3}$ .
Sum: $[f (t) + g (t)]^{'} = f^{'} (t) + g^{'} (t)$ : the derivative of a sum is the sum of the derivatives. For example $(t^{2} + 3 t)^{'} = (t^{2})^{'} + (3 t)^{'} = 2 t + 3$ .
Product: $\frac{d}{d t} (f \cdot g) = f^{'} g + f g^{'}$ . For example $\frac{d}{d t} (\sin t \cdot 2 t^{3}) = \cos t \cdot 2 t^{3} + \sin t \cdot 6 t^{2}$ .
Quotient: $\frac{d}{d t} (\frac{f}{g}) = \frac{f^{'} g - f g^{'}}{g^{2}}$ . For example $\frac{d}{d t} (\frac{\sin t}{t^{2}}) = \frac{\cos t \cdot t^{2} - \sin t \cdot 2 t}{t^{4}}$ .
Chain rule: $\frac{d}{d t} f (g (t)) = f^{'} (g (t)) \cdot g^{'} (t)$ .

Example. $h (t) = \sqrt{\sin t}$ is $f (g (t))$ with $f (t) = \sqrt{t}$ and $g (t) = \sin t$ , so $h^{'} (t) = \frac{1}{2 \sqrt{\sin t}} \cdot \cos t .$ For $f (t) = \sin (\ln t)$ : $\frac{d f}{d t} = \cos (\ln t) \cdot \frac{1}{t}$ .

Basic Integration Rules

An integral asks: which function has this derivative? In all of these, $c$ is a constant number.

$\int 1 d t = t + c, \int t d t = \frac{t^{2}}{2} + c, \int t^{n} d t = \frac{t^{n + 1}}{n + 1} + c (n \neq - 1),$

$\int \frac{1}{t} d t = \ln t + c, \int e^{t} d t = e^{t} + c, \int \sin t d t = - \cos t + c, \int \cos t d t = \sin t + c .$

Example. $\int t^{1 / 2} d t = \frac{t^{3 / 2}}{3 / 2} + c = \frac{2}{3} \sqrt{t^{3}} + c .$

The Definite Integral

Definition. The definite integral $\int_{a}^{b} f (t) d t$ is the area $A$ between the graph of $f$ and the $t$ -axis from $t = a$ to $t = b$ . In calculus it is defined as a limit of sums, $lim_{Δ t \to 0} \sum f (t) Δ t$ ; the integral sign $\int$ is a stretched S, for "sum", like the Greek capital sigma $Σ$ .

To compute it, find a function $F$ with $F^{'} = f$ and take $F (b) - F (a)$ , written $F (t) |_{a}^{b}$ .

Example. The area under $x = t$ from $t = 1$ to $t = 3$ is a trapezoid: $A = \frac{3 + 1}{2} \cdot 2 = 4$ . With the integral, $\int_{1}^{3} t d t = \frac{t^{2}}{2} |_{1}^{3} = \frac{3^{2}}{2} - \frac{1^{2}}{2} = \frac{9 - 1}{2} = 4.$ The area under $x = t^{2}$ from $1$ to $3$ is not a simple shape, but the integral still works: $\int_{1}^{3} t^{2} d t = \frac{t^{3}}{3} |_{1}^{3} = \frac{3^{3} - 1^{3}}{3} = \frac{26}{3} .$

The Derivative at a Point

The derivative of a function $f$ at a point $a$ is the number $f^{'} (a)$ .

Example. $x (t) = 3 t^{2} - \sin t$ . What is the derivative at $t = 1$ ? $\frac{d x}{d t} = 6 t - \cos t, {\frac{d x}{d t} |}_{t = 1} = 6 - \cos 1 \approx 5.46 .$

Example. The position of a particle at time $t$ is $x (t) = t^{3} - 2 t$ . Find its velocity at $t = 0$ and $t = 2$ , and its acceleration at $t = 2$ . $v (t) = 3 t^{2} - 2, v (0) = - 2, v (2) = 10,$ $a (t) = 6 t, a (2) = 12.$

Constant Acceleration, Seen Through Integration

For a constant acceleration $a (t) = a$ , integrating once gives $v (t) = a t + \dots$ and integrating again gives $x (t) = \frac{1}{2} a t^{2} + \dots$ . The degree goes up by one each time:

the $a$ – $t$ graph is a horizontal line (degree $0$ );
the $v$ – $t$ graph is a straight line (degree $1$ ), steeper for larger $| a |$ ;
the $x$ – $t$ graph is a parabola (degree $2$ ). When $a = 0$ the "parabola" is a straight line.

Example. A particle starts from rest at the position $0$ with acceleration $a (t) = t^{2} + \sin t$ .

(i) Find its velocity at $t = 2$ . (ii) Find its displacement between $t = 2$ and $t = 3$ .

(i) $v (t) = \frac{t^{3}}{3} - \cos t + c$ . From rest: $v (0) = 0 - \cos 0 + c = 0$ , so $c = 1$ and $v (t) = \frac{t^{3}}{3} - \cos t + 1, v (2) = \frac{8}{3} - \cos 2 + 1 \approx 4.08 m/s .$

(ii) $x (t) = \frac{t^{4}}{12} - \sin t + t + c$ . From $x (0) = 0 - \sin 0 + 0 + c = 0$ we get $c = 0$ , so $x (t) = \frac{t^{4}}{12} - \sin t + t$ and $x (3) - x (2) = (\frac{3^{4}}{12} - \sin 3 + 3) - (\frac{2^{4}}{12} - \sin 2 + 2) = \frac{65}{12} + 1 - \sin 3 + \sin 2 \approx 7.19 m .$

3. Free Fall: Vertical Motion

Vertical motion is the same as motion on a line, with two changes: we write $y (t)$ instead of $x (t)$ , with the positive direction up, and the acceleration is constant, $a = - g = - 9.81 {m/s}^{2} .$

Summary (Free fall). $y (t) = y_{0} + v_{0} t - \frac{1}{2} (9.81) t^{2}, v (t) = v_{0} - 9.81 t, a = - 9.81 .$

Remark. The acceleration of gravity always points down, towards the Earth. In the formulas the sign of $a$ is the sign of the direction we chose as positive: with "up is positive", $a = - g$ .

Example. A particle is thrown upward with the initial velocity $3$ m/s. Where is it at $t = 1$ s? $y = 0 + 3 \cdot 1 - \frac{1}{2} \cdot 9.81 \cdot 1^{2} \approx - 1.9 m .$ After one second it has already passed its starting point on the way down: it is $1.9$ m below it.

4. Vectors

In two dimensions velocity and acceleration are vectors with a direction in the plane, not only a sign. Before we move on to motion in the plane, we collect what we need about vectors.

Vectors and Scalars

Definition. A scalar is a quantity given by a single number, without direction: distance, speed, mass, time. A vector is a quantity with a direction and a magnitude (size): displacement, velocity, acceleration, force.

A vector is drawn as an arrow: its length is the magnitude and the way it points is the direction. Moving an arrow without turning it or changing its length does not change the vector. For example " $5$ units at the angle $θ$ " is the same vector wherever we draw it.

Definition. For the vector $\vec{R}$ from the origin to the point $(x, y)$ , the magnitude (length) is $| \vec{R} | = \sqrt{x^{2} + y^{2}},$ and its direction is the angle $θ$ with the positive $x$ -axis, where $\tan θ = \frac{y}{x}, θ = \tan^{- 1} (\frac{y}{x}) = \arctan (\frac{y}{x}) .$ The angle is measured counterclockwise (c.c.w.) from the positive $x$ -axis; a clockwise angle is negative.

The slope of the arrow is $\tan θ = \frac{y}{x}$ . In three dimensions there is a $z$ -axis too, and a vector is given by three numbers.

Remark. We name angles with Greek letters: $θ$ (theta), $α$ (alpha), $β$ (beta), $γ$ (gamma).

A Little Trigonometry

Left: a vector R from the origin to the point (x, y) at the angle theta, with components x = |R| cos theta and y = |R| sin theta. Right: a right triangle with the hypotenuse and the sides opposite and adjacent to theta — A vector and its two components form a right triangle.

Summary. In a right triangle with the angle $θ$ , $\sin θ = \frac{opposite}{hypotenuse}, \cos θ = \frac{adjacent}{hypotenuse}, \tan θ = \frac{opposite}{adjacent} .$

Degrees	$\sin$	$\cos$	$\tan$
$0^{\circ}$	$0$	$1$	$0$
$30^{\circ}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$
$45^{\circ}$	$\frac{1}{\sqrt{2}}$	$\frac{1}{\sqrt{2}}$	$1$
$60^{\circ}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$
$90^{\circ}$	$1$	$0$	undefined

These values come from two triangles: the $45^{\circ}$ – $45^{\circ}$ – $90^{\circ}$ triangle with sides $1$ , $1$ , $\sqrt{2}$ , and the $30^{\circ}$ – $60^{\circ}$ – $90^{\circ}$ triangle with sides $1$ , $\sqrt{3}$ , $2$ . A third triangle that appears often in physics problems is the $3$ – $4$ – $5$ triangle, whose angles are about $37^{\circ}$ and $53^{\circ}$ .

Definition. Angles are measured in degrees or in radians. A full turn is $360^{\circ} = 2 π$ rad, so $x^{\circ} = x \cdot \frac{π}{180} rad, θ rad = θ \cdot \frac{180}{π} degrees .$

Remark. Check which mode your calculator is in. If it works in radians, $\cos 30$ means the cosine of $30$ radians; for $30^{\circ}$ type $\cos (30 \cdot \frac{π}{180})$ .

Components of a Vector

Definition. The components of a vector $\vec{R}$ with magnitude $| \vec{R} |$ and direction $θ$ are $x = | \vec{R} | \cos θ, y = | \vec{R} | \sin θ .$ They tell how far the vector goes along each axis.

Example. A rope pulls a boat with the force $F = 5$ N at $30^{\circ}$ below the horizontal. The force pulls forward with $5 \cos 30^{\circ} \approx 4.33$ N and downward with $5 \sin 30^{\circ} = 2.5$ N.

On a slope of $37^{\circ}$ , a vector of length $5$ along the slope has the horizontal component $5 \cos 37^{\circ} \approx 4$ and the vertical component $5 \sin 37^{\circ} \approx 3$ .

The other way round, from the components we get the magnitude and the direction:

Example. $\vec{R}$ goes from the origin to $(5, 2)$ . Then $| \vec{R} | = \sqrt{5^{2} + 2^{2}} = \sqrt{29} \approx 5.385, θ = \arctan (\frac{2}{5}) \approx 0.38 rad = \frac{0.38 \cdot 180}{π} \approx {21.8}^{\circ} .$

Example. $\vec{R}$ goes from the origin to $(3, 4)$ . Then $| \vec{R} | = \sqrt{3^{2} + 4^{2}} = 5$ and $θ = \arctan (\frac{4}{3}) \approx {53.1}^{\circ}$ : the $3$ – $4$ – $5$ triangle.

Example. $\vec{R}$ goes from the origin to $(- 2, 2)$ . A calculator gives $\arctan (\frac{2}{- 2}) = \arctan (- 1) = - 45^{\circ}$ , but the vector points into the second quadrant (left and up). It makes $45^{\circ}$ with the negative $x$ -axis, so measured counterclockwise from the positive $x$ -axis its direction is $θ = 180^{\circ} - 45^{\circ} = 135^{\circ} .$

Remark. $\arctan$ only gives angles between $- 90^{\circ}$ and $90^{\circ}$ . When $x < 0$ (second or third quadrant), add $180^{\circ}$ to the calculator's answer. Always draw the vector first.

Adding Vectors

Definition. Tip to tail: to add $\vec{A} + \vec{B}$ , draw $\vec{B}$ starting at the tip of $\vec{A}$ . The sum, the resultant, goes from the tail of $\vec{A}$ to the tip of $\vec{B}$ .

Parallelogram: draw $\vec{A}$ and $\vec{B}$ from the same point. The sum is the diagonal of the parallelogram they span.

Left: B drawn from the tip of A, with A + B from the tail of A to the tip of B. Right: A and B from one point, with A + B the diagonal of their parallelogram — Both methods give the same resultant, and the order does not matter.

The parallelogram shows that $\vec{A} + \vec{B} = \vec{B} + \vec{A}$ . More vectors are added the same way, one after the other: for ${\vec{v}}_{1} + {\vec{v}}_{2} + {\vec{v}}_{3} + {\vec{v}}_{4}$ put each arrow at the tip of the one before, and the resultant goes from the tail of the first to the tip of the last.

Multiplying by a Number, Subtracting

A vector can be multiplied by a number (a scalar). $2 \vec{A}$ points the same way as $\vec{A}$ and is twice as long; $0.5 \vec{A}$ is half as long; $(- 1) \vec{A} = - \vec{A}$ has the same length and points the opposite way; $0 \cdot \vec{A}$ is the zero vector.

Definition. Subtraction is adding the opposite vector: $\vec{A} - \vec{B} = \vec{A} + (- \vec{B})$ .

In a drawing, if $\vec{A}$ and $\vec{B}$ start at the same point, $\vec{A} - \vec{B}$ goes from the tip of $\vec{B}$ to the tip of $\vec{A}$ . In the same way ${\vec{v}}_{1} - {\vec{v}}_{2} - {\vec{v}}_{3} + {\vec{v}}_{4}$ is drawn tip to tail with ${\vec{v}}_{2}$ and ${\vec{v}}_{3}$ reversed.

The Algebraic Method

In components, vectors are added and subtracted coordinate by coordinate. A vector that goes $5$ units in the $x$ direction and $3$ units in the $y$ direction is $(5, 3)$ .

Summary. If ${\vec{R}}_{1} = (x_{1}, y_{1})$ and ${\vec{R}}_{2} = (x_{2}, y_{2})$ , then ${\vec{R}}_{1} + {\vec{R}}_{2} = (x_{1} + x_{2}, y_{1} + y_{2}), c {\vec{R}}_{1} = (c x_{1}, c y_{1}) .$ The vector from the point $A$ to the point $B$ is $\vec{A B} = B - A$ .

Example. ${\vec{R}}_{1} = (1, 4)$ and ${\vec{R}}_{2} = (- 1, 5)$ give ${\vec{R}}_{1} + {\vec{R}}_{2} = (0, 9)$ .

From $A = (1, 4)$ to $B = (2, 8)$ : $\vec{A B} = (2, 8) - (1, 4) = (1, 4)$ , one unit right and four units up.

The Law of Sines and the Law of Cosines

When we add two vectors with a drawing, we get a triangle that need not have a right angle. Two rules connect its sides and angles. In a triangle $A B C$ , the side $a$ is opposite the angle $A$ , $b$ is opposite $B$ , and $c$ is opposite $C$ .

A triangle with the vertices A, B and C and the opposite sides a, b and c — Each side is named after the angle opposite it.

Theorem (Law of sines). $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} .$

Example. A triangle has the angles $10^{\circ}$ , $140^{\circ}$ and $30^{\circ}$ , and the side opposite the $10^{\circ}$ angle is $5$ m. Call $y$ the side opposite $140^{\circ}$ and $x$ the side opposite $30^{\circ}$ . Then $\frac{5}{\sin 10^{\circ}} = \frac{y}{\sin 140^{\circ}} = \frac{x}{\sin 30^{\circ}},$ $y = \frac{\sin 140^{\circ}}{\sin 10^{\circ}} \cdot 5 \approx 18.5 m, x = \frac{\sin 30^{\circ}}{\sin 10^{\circ}} \cdot 5 \approx 14.4 m .$

Theorem (Law of cosines). $c^{2} = a^{2} + b^{2} - 2 a b \cos C, a^{2} = b^{2} + c^{2} - 2 b c \cos A, b^{2} = a^{2} + c^{2} - 2 a c \cos B .$ For a right angle, $\cos 90^{\circ} = 0$ and this is the Pythagorean theorem.

Example. Two sides of a triangle are $8$ and $10$ , and the angle between them is $60^{\circ}$ . The third side $c$ satisfies $c^{2} = 8^{2} + 10^{2} - 2 \cdot 8 \cdot 10 \cos 60^{\circ} = 164 - 2 \cdot 8 \cdot 10 \cdot \frac{1}{2} = 84, c = \sqrt{84} = 2 \sqrt{21} \approx 9.17 .$

Adding Vectors with Components

The easiest way to add several vectors: split each into components, add the $x$ -components and the $y$ -components separately, and turn the result back into a magnitude and a direction.

Example. $| {\vec{R}}_{1} | = 10$ at $θ = 30^{\circ}$ , $| {\vec{R}}_{2} | = 3$ at $θ = 100^{\circ}$ , $| {\vec{R}}_{3} | = 9$ at $θ = 47^{\circ}$ . Find $\vec{R} = {\vec{R}}_{1} + {\vec{R}}_{2} + {\vec{R}}_{3}$ .

	$x$ -component	$y$ -component
${\vec{R}}_{1}$	$10 \cos 30^{\circ} \approx 8.66$	$10 \sin 30^{\circ} = 5$
${\vec{R}}_{2}$	$3 \cos 100^{\circ} \approx - 0.52$	$3 \sin 100^{\circ} \approx 2.95$
${\vec{R}}_{3}$	$9 \cos 47^{\circ} \approx 6.14$	$9 \sin 47^{\circ} \approx 6.58$
$\vec{R}$	$R_{x} \approx 14.28$	$R_{y} \approx 14.53$

So $\vec{R} \approx (14.28, 14.53)$ , with $| \vec{R} | = \sqrt{{14.28}^{2} + {14.53}^{2}} \approx 20.4, θ = \arctan (\frac{14.53}{14.28}) \approx {45.5}^{\circ} .$

Example. Two forces act on a point: $\vec{A}$ of $5$ N at $25^{\circ}$ and $\vec{B}$ of $10$ N at $135^{\circ}$ . Find $\vec{C} = \vec{A} + \vec{B}$ . $\vec{C} = (5 \cos 25^{\circ} + 10 \cos 135^{\circ}, 5 \sin 25^{\circ} + 10 \sin 135^{\circ}) \approx (- 2.54, 9.18),$ $| \vec{C} | = \sqrt{(- 2.54)^{2} + (9.18)^{2}} \approx 9.53 N .$ For the direction, $\arctan (\frac{9.18}{- 2.54}) \approx - {74.5}^{\circ}$ ; but $C_{x} < 0$ and $C_{y} > 0$ , so $\vec{C}$ is in the second quadrant and its angle is $- {74.5}^{\circ} + 180^{\circ} \approx {105.5}^{\circ}$ .

Summary and Strategy

Summary.

Vectors have a magnitude and a direction; scalars have only a magnitude. Force, displacement and velocity are vectors; distance, speed and mass are scalars.
A vector can be split into components along the axes.
Vectors are added tip to tail; the sum is called the resultant.
Two vectors in the same direction give a resultant whose magnitude is the sum of their magnitudes; two vectors in opposite directions give the difference of their magnitudes. Every other angle gives something in between.

When solving a vector problem:

Understand the frame of reference of the situation.
Choose a coordinate system that suits it. The axes need not be horizontal and vertical: for a block sliding down a slope, take the $x$ -axis along the slope.
For a drawing, pick a scale, for example $1$ cm for $1$ m.
Keep the directions given in the problem (north, east, south, west) the same in your drawing.
Connect the vectors tip to tail; the resultant goes from the tail of the first to the tip of the last.
Find the components. Most vector problems become simple in components.

Practice Problems

Exercise (1). A vector has the components $A_{x} = 2.5$ and $A_{y} = 7.5$ . What angle does it make with the positive $x$ -axis?

Solution. $θ = \arctan (\frac{7.5}{2.5}) = \arctan 3 \approx 1.249 rad = \frac{1.249 \cdot 180}{π} \approx 72^{\circ} .$

Exercise (2). Which pair of vectors can have a resultant of magnitude $35$ : $15$ and $15$ ; $20$ and $20$ ; $30$ and $70$ ; $20$ and $60$ ?

Solution. The resultant of two vectors lies between the difference and the sum of their magnitudes (opposite directions give the difference, the same direction the sum). The ranges are $0$ – $30$ , $0$ – $40$ , $40$ – $100$ and $40$ – $80$ . Only $20$ and $20$ can give $35$ .

Exercise (3). A vector of magnitude $17$ units makes $20^{\circ}$ with the positive $x$ -axis. What is its horizontal component?

Solution. $17 \cos 20^{\circ} \approx 16$ units.

Exercise (4). As the angle between a vector and the horizontal grows from $0^{\circ}$ to $90^{\circ}$ , what happens to its vertical component?

Solution. The vertical component is $| \vec{A} | \sin θ$ , and $\sin θ$ grows from $0$ to $1$ on this range, so the vertical component increases.

Exercise (5). Which of these sets of displacements have the same resultant? I. $6$ m east, $9$ m north, $12$ m west. II. $6$ m north, $9$ m west, $12$ m east. III. $6$ m east, $12$ m west, $9$ m north. IV. $9$ m north, $6$ m east, $12$ m west.

Solution. Take east as $+ x$ and north as $+ y$ and add components. I: $(6 - 12, 9) = (- 6, 9)$ . II: $(- 9 + 12, 6) = (3, 6)$ . III: $(6 - 12, 9) = (- 6, 9)$ . IV: $(6 - 12, 9) = (- 6, 9)$ . So I, III and IV have the same resultant; the order of the steps does not matter.

Vector Notation

Definition. $\hat{ı}$ and $\hat{ȷ}$ are the vectors of length $1$ in the positive $x$ and $y$ directions. Every vector in the plane can be written with them: $\vec{r} = x \hat{ı} + y \hat{ȷ} .$ In three dimensions a third unit vector $\hat{k}$ is added for the $z$ -axis.

For example $\vec{v} = 4.5 \hat{ı} + 4 \hat{ȷ}$ goes $4.5$ units to the right and $4$ units up. Vectors are added and multiplied by numbers component by component.

Example. Let ${\vec{r}}_{1} = 2 \hat{ı} + 3 \hat{ȷ}$ and ${\vec{r}}_{2} = 1.5 \hat{ı} - 4.2 \hat{ȷ}$ . Then ${\vec{r}}_{1} + {\vec{r}}_{2} = 3.5 \hat{ı} - 1.2 \hat{ȷ}, 2 {\vec{r}}_{1} = 4 \hat{ı} + 6 \hat{ȷ}, (- 1) {\vec{r}}_{1} = - 2 \hat{ı} - 3 \hat{ȷ} .$

5. Kinematics in Two Dimensions

Position, Velocity and Acceleration in Two Dimensions

Definition. The position of a particle moving in the plane is given by two component functions: $\vec{r} (t) = x (t) \hat{ı} + y (t) \hat{ȷ} .$ The curve that the tip of $\vec{r} (t)$ draws is the trajectory of the particle.

How do we find the velocity vector at any time $t$ ? We differentiate each component.

Summary. $\vec{v} (t) = \frac{d \vec{r}}{d t} = \frac{d x}{d t} \hat{ı} + \frac{d y}{d t} \hat{ȷ} = v_{x} \hat{ı} + v_{y} \hat{ȷ},$ $\vec{a} (t) = \frac{d \vec{v}}{d t} = \frac{d^{2} x}{d t^{2}} \hat{ı} + \frac{d^{2} y}{d t^{2}} \hat{ȷ} = a_{x} \hat{ı} + a_{y} \hat{ȷ}, a_{x} = \frac{d v_{x}}{d t}, a_{y} = \frac{d v_{y}}{d t} .$ The velocity vector is tangent to the trajectory.

Example. $\vec{r} (t) = t^{2} \hat{ı} - 4 \sin t \hat{ȷ}$ . Then $\vec{v} (t) = 2 t \hat{ı} - 4 \cos t \hat{ȷ}$ . At $t = 2$ s, $\vec{r} (2) = 4 \hat{ı} - 4 \sin 2 \hat{ȷ} \approx 4 \hat{ı} - 3.637 \hat{ȷ},$ $\vec{v} (2) = 4 \hat{ı} - 4 \cos 2 \hat{ȷ} \approx 4 \hat{ı} + 1.665 \hat{ȷ} .$

The Complete Definitions

Summary.

Instantaneous velocity: $\vec{v} (t) = lim_{Δ t \to 0} \frac{Δ \vec{r}}{Δ t} = \frac{d \vec{r}}{d t} ⟺ Δ \vec{r} = \int d \vec{r} = \int_{t_{1}}^{t_{2}} \vec{v} (t) d t$ . Here $Δ \vec{r}$ is the change of position, the displacement.
Instantaneous speed: $| \vec{v} | = \sqrt{v_{x}^{2} + v_{y}^{2}}$ , the magnitude of the velocity vector.
Instantaneous acceleration: $\vec{a} (t) = lim_{Δ t \to 0} \frac{Δ \vec{v}}{Δ t} = \frac{d \vec{v}}{d t} = \frac{d^{2} \vec{r}}{d t^{2}} ⟺ Δ \vec{v} = \int d \vec{v} = \int_{t_{1}}^{t_{2}} \vec{a} (t) d t$ . Here $Δ \vec{v}$ is the change of velocity.

Remark. In two dimensions $Δ \vec{r}$ , $Δ \vec{v}$ and $Δ \vec{a}$ are all vectors.

The Displacement Vector

Definition. If the particle is at ${\vec{r}}_{1} = \vec{r} (t_{1})$ at time $t_{1}$ and at ${\vec{r}}_{2} = \vec{r} (t_{2})$ at time $t_{2}$ , its displacement vector is $Δ \vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1} .$ It depends only on the two end points, not on the path between them.

Example. If ${\vec{r}}_{1} = 2 \hat{ı} - 5 \hat{ȷ}$ and ${\vec{r}}_{2} = 3 \hat{ı} + 4 \hat{ȷ}$ , then $Δ \vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1} = \hat{ı} + 9 \hat{ȷ}, | Δ \vec{r} | = \sqrt{1 + 81} = \sqrt{82} \approx 9.06 .$

Example. $\vec{r} (t) = 2 t \hat{ı} - 3 t^{2} \hat{ȷ}$ . What is the displacement vector between $t = 1$ and $t = 3$ ? $\vec{r} (1) = 2 \hat{ı} - 3 \hat{ȷ}, \vec{r} (3) = 6 \hat{ı} - 27 \hat{ȷ}, Δ \vec{r} = \vec{r} (3) - \vec{r} (1) = 4 \hat{ı} - 24 \hat{ȷ} .$ The velocity and the speed are $\vec{v} (t) = 2 \hat{ı} - 6 t \hat{ȷ}, | \vec{v} (t) | = \sqrt{2^{2} + (6 t)^{2}} = \sqrt{36 t^{2} + 4},$ and the acceleration is $\vec{a} (t) = - 6 \hat{ȷ}$ . Integrating the speed gives the distance travelled along the path; in the first three seconds it is $\int_{0}^{3} \sqrt{36 t^{2} + 4} d t \approx 28.1 m$ (by calculator), more than the length of any straight segment between the end points.

Constant Acceleration in Two Dimensions

The formulas pass from one dimension to two by putting vectors in place of numbers:

Summary. $\vec{r} (t) = {\vec{r}}_{0} + {\vec{v}}_{0} t + \frac{1}{2} \vec{a} t^{2}, \vec{v} (t) = {\vec{v}}_{0} + \vec{a} t .$ We work with the component functions separately: $x (t) = x_{0} + v_{x 0} t + \frac{1}{2} a_{x} t^{2}, y (t) = y_{0} + v_{y 0} t + \frac{1}{2} a_{y} t^{2},$ $v_{x} (t) = v_{x 0} + a_{x} t, v_{y} (t) = v_{y 0} + a_{y} t .$

Example. A particle is at $(0, 0)$ at $t = 0$ and has the velocity $\vec{v} (t) = - 3 t \hat{ı} + 4 \hat{ȷ}$ .

(i) Find $\vec{a}$ . (ii) Find the position function $\vec{r} (t)$ .

(i) $\vec{a} (t) = \frac{d \vec{v}}{d t} = - 3 \hat{ı}$ .

(ii) Integrate each component ( $t^{1}$ becomes $\frac{t^{2}}{2}$ ): $\vec{r} (t) = (- \frac{3}{2} t^{2} + c_{1}) \hat{ı} + (4 t + c_{2}) \hat{ȷ} .$ From $\vec{r} (0) = c_{1} \hat{ı} + c_{2} \hat{ȷ} = (0, 0)$ we get $c_{1} = c_{2} = 0$ , so $\vec{r} (t) = - \frac{3}{2} t^{2} \hat{ı} + 4 t \hat{ȷ} .$

The Dot Product

Definition. The dot product (scalar product) of $\vec{u} = x_{1} \hat{ı} + y_{1} \hat{ȷ}$ and $\vec{v} = x_{2} \hat{ı} + y_{2} \hat{ȷ}$ is the number $\vec{u} \cdot \vec{v} = x_{1} x_{2} + y_{1} y_{2} .$

Theorem. $\vec{u} \cdot \vec{v} = | \vec{u} | | \vec{v} | \cos θ$ , where $| \vec{u} |$ and $| \vec{v} |$ are the lengths of the vectors and $θ$ is the angle between them. So $\cos θ = \frac{\vec{u} \cdot \vec{v}}{| \vec{u} | | \vec{v} |} .$

Example: Two Paths That Meet

Example. Sally is chasing Calvin. Calvin moves along the path ${\vec{r}}_{C} (t) = (t - 1) \hat{ı} + (11 - t^{2}) \hat{ȷ}$ and Sally along ${\vec{r}}_{S} (t) = (5 - 5 t) \hat{ı} + (5 + 5 t) \hat{ȷ}$ .

(a) Will Sally catch Calvin? (b) If so, what is the angle between their velocities when they meet?

(a) They meet at a time $t_{1}$ when ${\vec{r}}_{C} (t_{1}) = {\vec{r}}_{S} (t_{1})$ , that is, when both coordinates are equal at the same moment. The $x$ -coordinates are equal when $t_{1} - 1 = 5 - 5 t_{1} \Rightarrow 6 t_{1} = 6 \Rightarrow t_{1} = 1.$ This is the only time at which they have the same $x$ -coordinate. At that moment the $y$ -coordinates are $y_{C} = 11 - 1 = 10$ and $y_{S} = 5 + 5 = 10$ , also equal. So yes: at $t = 1$ they meet at the point $(0, 10)$ .

(b) The velocities are ${\vec{v}}_{C} (t) = \frac{d {\vec{r}}_{C}}{d t} = \hat{ı} - 2 t \hat{ȷ}, {\vec{v}}_{S} (t) = \frac{d {\vec{r}}_{S}}{d t} = - 5 \hat{ı} + 5 \hat{ȷ},$ so at $t = 1$ : ${\vec{v}}_{C} (1) = \hat{ı} - 2 \hat{ȷ}$ and ${\vec{v}}_{S} (1) = - 5 \hat{ı} + 5 \hat{ȷ}$ . Then $\cos θ = \frac{(1) (- 5) + (- 2) (5)}{\sqrt{1^{2} + (- 2)^{2}} \sqrt{(- 5)^{2} + 5^{2}}} = \frac{- 15}{\sqrt{5} \cdot 5 \sqrt{2}} \approx - 0.949, θ = \cos^{- 1} (- 0.949) \approx {161.6}^{\circ} .$

The same angle can be read from a drawing: ${\vec{v}}_{S}$ makes $135^{\circ}$ with the positive $x$ -axis, and ${\vec{v}}_{C}$ points $\tan^{- 1} 2 \approx {63.4}^{\circ}$ below it, so the angle between them is $135^{\circ} + (90^{\circ} - {63.4}^{\circ}) \approx {161.6}^{\circ}$ .

Example: From Velocity to Acceleration and Displacement

Example. A particle moves in the plane with the velocity $\vec{v} (t) = 4 t^{3} \hat{ı} - \cos t \hat{ȷ}$ (in m/s).

(a) What is its acceleration as a function of time? (b) What is its displacement between $t = 0$ and $t = 3$ ?

(a) $\vec{a} (t) = \frac{d \vec{v}}{d t} = 12 t^{2} \hat{ı} + \sin t \hat{ȷ}$ .

(b) We need a function $\vec{r} (t) = x (t) \hat{ı} + y (t) \hat{ȷ}$ whose derivative is $\vec{v}$ : $\vec{r} (t) = (t^{4} + c_{1}) \hat{ı} + (- \sin t + c_{2}) \hat{ȷ} .$ The derivative of $t^{4} + c_{1}$ is $4 t^{3}$ for every constant $c_{1}$ ; in the same way the derivative of $- \sin t + c_{2}$ is $- \cos t$ for every $c_{2}$ . The constants cancel in the displacement: $\vec{r} (3) - \vec{r} (0) = [(81 + c_{1}) \hat{ı} + (- \sin 3 + c_{2}) \hat{ȷ}] - [(0 + c_{1}) \hat{ı} + (0 + c_{2}) \hat{ȷ}] = 81 \hat{ı} - \sin 3 \hat{ȷ} \approx 81 \hat{ı} - 0.141 \hat{ȷ} .$ ( $\sin 3$ is in radians.)

6. Projectile Motion

Polar and Rectangular Coordinates

Every point of the plane can be given in two ways: by its rectangular coordinates $(x, y)$ , or by its polar coordinates $(r, θ)$ , the distance to the origin $r \geq 0$ and the angle of rotation $θ$ from the positive $x$ -axis.

Summary. $polar \to rectangular: x = r \cos θ, y = r \sin θ,$ $rectangular \to polar: r = \sqrt{x^{2} + y^{2}}, θ = \arctan (\frac{y}{x}) .$

For a velocity $\vec{v} (t) = v_{x} (t) \hat{ı} + v_{y} (t) \hat{ȷ}$ the same formulas give the speed and the direction of motion: $| \vec{v} | = \sqrt{v_{x}^{2} + v_{y}^{2}}, θ = \tan^{- 1} (\frac{v_{y}}{v_{x}}), v_{x} = | \vec{v} | \cos θ, v_{y} = | \vec{v} | \sin θ .$

Example. A particle is thrown with the speed $5$ m/s at an angle of $60^{\circ}$ above the horizontal. Its velocity components are $v_{x} = 5 \cos 60^{\circ} = 2.5 m/s, v_{y} = 5 \sin 60^{\circ} \approx 4.33 m/s .$

The Equations of Projectile Motion

Note (Standard assumptions). Unless told otherwise, there is no air resistance, and the only force on the object is gravity, which gives it the downward acceleration $g = 9.8 {m/s}^{2}$ (we often use $9.81$ , or $10$ for easy numbers).

We analyse projectile motion as two one-dimensional motions, one on the $x$ -axis and one on the $y$ -axis, superposed. A projectile launched with speed $v_{0}$ at the angle $θ$ has $v_{x 0} = v_{0} \cos θ$ and $v_{y 0} = v_{0} \sin θ$ .

Summary (Projectile motion). On the $x$ -axis there is no acceleration: $a_{x} = 0, v_{x} (t) = v_{x 0}, x (t) = x_{0} + v_{x 0} t, v_{x 0} = v_{0} \cos θ .$ On the $y$ -axis the acceleration is $- g$ : $a_{y} = - g, v_{y} = v_{y 0} - g t, y = y_{0} + v_{y 0} t - \frac{1}{2} g t^{2}, v_{y}^{2} = v_{y 0}^{2} - 2 g (y - y_{0}), v_{y 0} = v_{0} \sin θ .$

Here $v_{x} (t)$ is the horizontal component of the velocity vector at time $t$ ; it never changes.

Problem-Solving Tips

At the peak of the path, $v_{y} = 0$ .
Because of the symmetry of the path, at the same height on the way down the speed is the same as on the way up, and the velocity makes the same angle $θ$ with the horizontal, now below it: $v_{x}$ is unchanged and $v_{y}$ has become $- v_{y 0}$ .
We work on the motion in two dimensions, $x$ and $y$ , separately; the time $t$ is what connects them.

A projectile path with velocity components drawn at the launch point, at the peak and at the same height on the way down — The horizontal velocity never changes; the vertical velocity falls to $0$ at the peak and returns with the opposite sign.

The Trajectory Is a Parabola

Exercise. Show that $y (x)$ , the trajectory of a projectile, is a parabola.

What does "parabolic" mean? It means that the relation between $y$ and $x$ is like $y = x^{2}$ : $y$ is a polynomial of degree $2$ in $x$ . For example $y = x^{2} + 2 x = (x + 1)^{2} - 1$ is the parabola $y = x^{2}$ shifted one unit to the left and one unit down.

Solution. We may choose the coordinates as we wish; let the projectile start from $(0, 0)$ . Then $x (t) = v_{x} t, y (t) = v_{y 0} t - \frac{1}{2} g t^{2} .$ From the first equation $t = \frac{x}{v_{x}}$ . Putting this into the second, $y (x) = v_{y 0} \frac{x}{v_{x}} - \frac{1}{2} g \frac{x^{2}}{v_{x}^{2}} = (\frac{v_{y 0}}{v_{x}}) x + (\frac{- g}{2 v_{x}^{2}}) x^{2},$ a polynomial of degree $2$ in $x$ : the trajectory is a parabola, opening downward.

If we write $v_{y 0}$ with the speed, $v_{y 0} = \sqrt{v^{2} - v_{x}^{2}}$ , the same result reads $y (x) = x \sqrt{\frac{v^{2}}{v_{x}^{2}} - 1} - \frac{g}{2 v_{x}^{2}} x^{2}$ . $◼$

Example: Thrown off a Cliff

Example. An object is thrown off a cliff $50$ m high with the initial velocity $25$ m/s at $30^{\circ}$ above the horizontal. How far from the base of the cliff does it land? (Take $g = 10 {m/s}^{2}$ .)

Put the origin at the top of the cliff, where the object is thrown. Then $v_{x 0} = 25 \cos 30^{\circ} \approx 21.65 m/s, v_{y 0} = 25 \sin 30^{\circ} = 12.5 m/s .$ The distance is $d = v_{x 0} t$ , where $t$ is the total time of the motion. The object lands when $y = - 50$ : $- 50 = 0 + 12.5 t - \frac{1}{2} \cdot 10 t^{2} \Rightarrow 5 t^{2} - 12.5 t - 50 = 0.$ $Δ = (- 12.5)^{2} - 4 (5) (- 50) = 1156.25, t = \frac{12.5 \pm \sqrt{1156.25}}{2 \cdot 5} = \frac{12.5 \pm 34}{10} .$ The negative root has no meaning here, so $t = 4.65$ s and $d = v_{x 0} t \approx 21.65 \times 4.65 \approx 101 m .$

A ball thrown from the top of a 50 m cliff at 25 m/s, 30 degrees above the horizontal, landing about 101 m from the base — The time comes from the vertical motion; the distance from the horizontal one.

Example: The Monkey and the Dart

Example. A naturalist wants to shoot a monkey with a tranquilizer dart. She aims the dart directly at the monkey. At the moment she shoots, the monkey lets go of the branch and falls freely from rest. Prove that the dart hits the monkey (assuming the dart reaches the monkey's $x$ -position before either of them hits the ground).

The dart leaves along the line of sight to the monkey; both the dart and the monkey fall one half g t squared below where they would be without gravity, so they meet — Gravity pulls the dart and the monkey down by the same amount, $\frac{1}{2} g t^{2}$ .

Proof. Put the gun at the origin, let the monkey start at $(X, Y)$ , and let the dart leave with speed $v$ at the angle $θ$ . The dart is aimed at the monkey, so $\tan θ = \frac{Y}{X} .$ The dart reaches the monkey's $x$ -position $X$ at the time $t = \frac{X}{v \cos θ} .$ At that moment the height of the dart is $b = (v \sin θ) (\frac{X}{v \cos θ}) - \frac{1}{2} g {(\frac{X}{v \cos θ})}^{2} = X \tan θ - \frac{1}{2} g t^{2} = Y - \frac{1}{2} g t^{2},$ and the monkey, falling from rest, is at the height $Y - \frac{1}{2} g t^{2} .$ The two heights are equal, so the dart hits the monkey. Without gravity the dart would fly along the line of sight and the monkey would stay put; gravity moves both of them down by the same $\frac{1}{2} g t^{2}$ . $◼$

Flashcards

One card for every definition in these notes. Tap or click a card (or press the space bar) to see the definition; use the arrows to move between cards.

Velocity $v (t)$

The derivative of the position: $v (t) = \frac{d x}{d t} = lim_{Δ t \to 0} \frac{Δ x}{Δ t}$ , the slope of the tangent line to the $x$ – $t$ graph. Its sign gives the direction of motion.

Average velocity

$\bar{v} = \frac{Δ x}{Δ t} = \frac{x_{2} - x_{1}}{t_{2} - t_{1}}$ , the slope of the secant line between two points of the $x$ – $t$ graph.

Acceleration $a (t)$

The derivative of the velocity: $a (t) = \frac{d v}{d t} = \frac{d^{2} x}{d t^{2}}$ . The average acceleration is $\bar{a} = \frac{Δ v}{Δ t}$ .

Speed

The magnitude of the velocity: $| v (t) |$ in one dimension, $| \vec{v} | = \sqrt{v_{x}^{2} + v_{y}^{2}}$ in two dimensions.

Average speed

$\frac{total distance}{Δ t}$ . It can differ from the size of the average velocity, because distance counts every part of the path and displacement only the end points.

Displacement

The change of position: $Δ x = x_{2} - x_{1} = \int_{t_{1}}^{t_{2}} v (t) d t$ on a line, $Δ \vec{r} = {\vec{r}}_{2} - {\vec{r}}_{1}$ in the plane.

Definite integral $\int_{a}^{b} f (t) d t$

The area between the graph of $f$ and the $t$ -axis from $a$ to $b$ (area below the axis counts as negative). It equals $F (b) - F (a)$ for any $F$ with $F^{'} = f$ .

Scalar and vector

A scalar is a single number without direction (distance, speed, mass, time). A vector has a magnitude and a direction (displacement, velocity, acceleration, force).

Radian

A unit of angle: a full turn is $360^{\circ} = 2 π$ rad, so $x^{\circ} = x \cdot \frac{π}{180}$ rad.

Components of a vector

For magnitude $| \vec{R} |$ and direction $θ$ : $x = | \vec{R} | \cos θ$ and $y = | \vec{R} | \sin θ$ , how far the vector goes along each axis.

Resultant (tip to tail, parallelogram)

The sum of vectors. Tip to tail: draw $\vec{B}$ from the tip of $\vec{A}$ ; $\vec{A} + \vec{B}$ goes from the tail of $\vec{A}$ to the tip of $\vec{B}$ . Parallelogram: draw both from one point; the sum is the diagonal.

Vector subtraction $\vec{A} - \vec{B}$

Adding the opposite vector: $\vec{A} - \vec{B} = \vec{A} + (- \vec{B})$ , where $- \vec{B}$ has the length of $\vec{B}$ and points the opposite way.

Law of sines

In a triangle with sides $a, b, c$ opposite the angles $A, B, C$ : $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$ .

Law of cosines

$c^{2} = a^{2} + b^{2} - 2 a b \cos C$ , where $C$ is the angle between the sides $a$ and $b$ . For $C = 90^{\circ}$ it is the Pythagorean theorem.

Magnitude and direction of a vector $(x, y)$

Magnitude $\sqrt{x^{2} + y^{2}}$ ; direction the angle $θ$ measured counterclockwise from the positive $x$ -axis, $\tan θ = \frac{y}{x}$ (add $180^{\circ}$ to the calculator's $\arctan$ when $x < 0$ ).

Unit vectors $\hat{ı}$ , $\hat{ȷ}$

The vectors of length $1$ in the positive $x$ and $y$ directions; every vector in the plane is $x \hat{ı} + y \hat{ȷ}$ .

Position vector $\vec{r} (t)$ and trajectory

$\vec{r} (t) = x (t) \hat{ı} + y (t) \hat{ȷ}$ , given by two component functions. The curve its tip draws is the trajectory.

Dot product $\vec{u} \cdot \vec{v}$

$x_{1} x_{2} + y_{1} y_{2}$ for $\vec{u} = (x_{1}, y_{1})$ , $\vec{v} = (x_{2}, y_{2})$ . It equals $| \vec{u} | | \vec{v} | \cos θ$ , so $\cos θ = \frac{\vec{u} \cdot \vec{v}}{| \vec{u} | | \vec{v} |}$ .

Polar coordinates $(r, θ)$

The distance $r \geq 0$ to the origin and the angle $θ$ from the positive $x$ -axis: $x = r \cos θ$ , $y = r \sin θ$ , $r = \sqrt{x^{2} + y^{2}}$ .

AP Physics C Mechanics: Lecture Notes

1. Kinematics in One Dimension

Position, Velocity and Acceleration

Average and Instantaneous Velocity

The Power Rule

Acceleration

Derivative and Integral: Reading the Graphs

General Formulas

Derivative Examples

Equations of Motion with Constant Acceleration

Graphs of x, v and a

Practice Problems

2. A Little Calculus: Derivatives and Integrals

Derivatives of Basic Functions

Examples of the Power Rule

Basic Rules of Differentiation

Basic Integration Rules

The Definite Integral

The Derivative at a Point

Constant Acceleration, Seen Through Integration

3. Free Fall: Vertical Motion

4. Vectors

Vectors and Scalars

A Little Trigonometry

Components of a Vector

Adding Vectors

Multiplying by a Number, Subtracting

The Algebraic Method

The Law of Sines and the Law of Cosines

Adding Vectors with Components

Summary and Strategy

Practice Problems

Vector Notation

5. Kinematics in Two Dimensions

Position, Velocity and Acceleration in Two Dimensions

The Complete Definitions

The Displacement Vector

Constant Acceleration in Two Dimensions

The Dot Product

Example: Two Paths That Meet

Example: From Velocity to Acceleration and Displacement

6. Projectile Motion

Polar and Rectangular Coordinates

The Equations of Projectile Motion

Problem-Solving Tips

The Trajectory Is a Parabola

Example: Thrown off a Cliff

Example: The Monkey and the Dart

Flashcards

Graphs of $x$ , $v$ and $a$