College Physics for AP® Courses

# Section Summary

### 3.1Kinematics in Two Dimensions: An Introduction

• The shortest path between any two points is a straight line. In two dimensions, this path can be represented by a vector with horizontal and vertical components.
• The horizontal and vertical components of a vector are independent of one another. Motion in the horizontal direction does not affect motion in the vertical direction, and vice versa.

### 3.2Vector Addition and Subtraction: Graphical Methods

• The graphical method of adding vectors $AA size 12{A} {}$ and $BB size 12{B} {}$ involves drawing vectors on a graph and adding them using the head-to-tail method. The resultant vector $RR size 12{A} {}$ is defined such that $A+B=RA+B=R$. The magnitude and direction of $RR size 12{A} {}$ are then determined with a ruler and protractor, respectively.
• The graphical method of subtracting vector $B B$ from $AA$ involves adding the opposite of vector $BB$, which is defined as $−B−B size 12{ - B} {}$. In this case, $A–B=A+(–B)=RA–B=A+(–B)=R$. Then, the head-to-tail method of addition is followed in the usual way to obtain the resultant vector $RR$.
• Addition of vectors is commutative such that .
• The head-to-tail method of adding vectors involves drawing the first vector on a graph and then placing the tail of each subsequent vector at the head of the previous vector. The resultant vector is then drawn from the tail of the first vector to the head of the final vector.
• If a vector $AA size 12{A} {}$ is multiplied by a scalar quantity $cc size 12{A} {}$, the magnitude of the product is given by $cAcA size 12{ ital "cA"} {}$. If $cc size 12{c} {}$ is positive, the direction of the product points in the same direction as $AA size 12{A} {}$; if $cc size 12{c} {}$ is negative, the direction of the product points in the opposite direction as $AA size 12{A} {}$.

### 3.3Vector Addition and Subtraction: Analytical Methods

• The analytical method of vector addition and subtraction involves using the Pythagorean theorem and trigonometric identities to determine the magnitude and direction of a resultant vector.
• The steps to add vectors $AA size 12{A} {}$ and $BB size 12{B} {}$ using the analytical method are as follows:

Step 1: Determine the coordinate system for the vectors. Then, determine the horizontal and vertical components of each vector using the equations

A x = A cos θ B x = B cos θ A x = A cos θ B x = B cos θ alignl { stack { size 12{A rSub { size 8{x} } =A"cos"θ} {} # B rSub { size 8{x} } =B"cos"θ {} } } {}

and

A y = A sin θ By = B sin θ . A y = A sin θ By = B sin θ . alignl { stack { size 12{A rSub { size 8{y} } =A" sin"θ} {} # B=B suby " sin "θ {} } } {}

Step 2: Add the horizontal and vertical components of each vector to determine the components $RxRx size 12{R rSub { size 8{x} } } {}$ and $RyRy size 12{R rSub { size 8{y} } } {}$ of the resultant vector, $RR size 12{R} {}$:

$R x = A x + B x R x = A x + B x size 12{R rSub { size 8{x} } =A rSub { size 8{x} } +B rSub { size 8{x} } } {}$

and

$Ry=Ay+By.Ry=Ay+By. size 12{R rSub { size 8{y} } =A rSub { size 8{y} } +B rSub { size 8{y} } } {}$

Step 3: Use the Pythagorean theorem to determine the magnitude, $RR size 12{R} {}$, of the resultant vector $RR size 12{R} {}$:

$R=Rx2+Ry2.R=Rx2+Ry2. size 12{R= sqrt {R rSub { size 8{x} } rSup { size 8{2} } +R rSub { size 8{y} } rSup { size 8{2} } } } {}$

Step 4: Use a trigonometric identity to determine the direction, $θθ size 12{θ} {}$, of $RR size 12{R} {}$:

$θ=tan−1(Ry/Rx).θ=tan−1(Ry/Rx). size 12{θ="tan" rSup { size 8{ - 1} } $$R rSub { size 8{y} } /R rSub { size 8{x} }$$ } {}$

### 3.4Projectile Motion

• Projectile motion is the motion of an object through the air that is subject only to the acceleration of gravity.
• To solve projectile motion problems, perform the following steps:
1. Determine a coordinate system. Then, resolve the position and/or velocity of the object in the horizontal and vertical components. The components of position $ss size 12{s} {}$ are given by the quantities $xx size 12{x} {}$ and $yy size 12{y} {}$, and the components of the velocity $vv size 12{v} {}$ are given by $vx=vcosθvx=vcosθ size 12{v rSub { size 8{x} } =v"cos"θ} {}$ and $vy=vsinθvy=vsinθ size 12{v rSub { size 8{y} } =v"sin"θ} {}$, where $vv size 12{v} {}$ is the magnitude of the velocity and $θθ size 12{θ} {}$ is its direction.
2. Analyze the motion of the projectile in the horizontal direction using the following equations:
$Horizontal motion ( a x = 0 ) Horizontal motion ( a x = 0 ) size 12{"Horizontal motion " $$a rSub { size 8{x} } =0$$ } {}$
$x = x 0 + v x t x = x 0 + v x t size 12{x=x rSub { size 8{0} } +v rSub { size 8{x} } t} {}$
$vx=v0x= v x =velocity is a constant.vx=v0x= v x =velocity is a constant. size 12{v rSub { size 8{x} } =v rSub { size 8{0x} } =v rSub { size 8{x} } ="velocity is a constant."} {}$
3. Analyze the motion of the projectile in the vertical direction using the following equations:
$Vertical motion ( Assuming positive direction is up; a y = − g = − 9 . 80 m /s 2 ) Vertical motion ( Assuming positive direction is up; a y = − g = − 9 . 80 m /s 2 ) size 12{"Vertical motion " $$"Assuming positive direction is up; "a rSub { size 8{y} } = - g= - 9 "." "80"" m/s" rSup { size 8{2} }$$ } {}$
$y = y 0 + 1 2 ( v 0y + v y ) t y = y 0 + 1 2 ( v 0y + v y ) t size 12{y=y rSub { size 8{0} } + { {1} over {2} } $$v rSub { size 8{0y} } +v rSub { size 8{y} }$$ t} {}$
$v y = v 0 y − gt v y = v 0 y − gt size 12{v rSub { size 8{y} } =v rSub { size 8{0y} } - ital "gt"} {}$
$y = y 0 + v 0 y t − 1 2 gt 2 y = y 0 + v 0 y t − 1 2 gt 2 size 12{y=y rSub { size 8{0} } +v rSub { size 8{0y} } t - { {1} over {2} } ital "gt" rSup { size 8{2} } } {}$
$vy2=v0y2−2g(y−y0).vy2=v0y2−2g(y−y0). size 12{v rSub { size 8{y} } rSup { size 8{2} } =v rSub { size 8{0y} } rSup { size 8{2} } - 2g $$y - y rSub { size 8{0} }$$ } {}$
4. Recombine the horizontal and vertical components of location and/or velocity using the following equations:
$s = x 2 + y 2 s = x 2 + y 2 size 12{s= sqrt {x rSup { size 8{2} } +y rSup { size 8{2} } } } {}$
$θ = tan − 1 ( y / x ) θ = tan − 1 ( y / x ) size 12{θ="tan" rSup { size 8{ - 1} } $$y/x$$ } {}$
$v = v x 2 + v y 2 v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}$
$θv=tan−1(vy/vx).θv=tan−1(vy/vx). size 12{θ rSub { size 8{v} } ="tan" rSup { size 8{ - 1} } $$v rSub { size 8{y} } /v rSub { size 8{x} }$$ } {}$
• The maximum height $hh size 12{h} {}$ of a projectile launched with initial vertical velocity $v0yv0y size 12{v rSub { size 8{0y} } } {}$ is given by
$h=v0y22g.h=v0y22g. size 12{h= { {v rSub { size 8{0y} } rSup { size 8{2} } } over {2g} } } {}$
• The maximum horizontal distance traveled by a projectile is called the range. The range $RR size 12{R} {}$ of a projectile on level ground launched at an angle $θ0θ0 size 12{θ rSub { size 8{0} } } {}$ above the horizontal with initial speed $v0v0 size 12{v rSub { size 8{0} } } {}$ is given by
$R=v02sin2θ0g.R=v02sin2θ0g. size 12{R= { {v rSub { size 8{0} } rSup { size 8{2} } "sin"2θ rSub { size 8{0} } } over {g} } } {}$

• Velocities in two dimensions are added using the same analytical vector techniques, which are rewritten as
$v x = v cos θ v x = v cos θ size 12{v rSub { size 8{x} } =v"cos"θ} {}$
$v y = v sin θ v y = v sin θ size 12{v rSub { size 8{y} } =v"sin"θ} {}$
$v = v x 2 + v y 2 v = v x 2 + v y 2 size 12{v= sqrt {v rSub { size 8{x} } rSup { size 8{2} } +v rSub { size 8{y} } rSup { size 8{2} } } } {}$
$θ=tan−1(vy/vx).θ=tan−1(vy/vx). size 12{θ="tan" rSup { size 8{ - 1} } $$v rSub { size 8{y} } /v rSub { size 8{x} }$$ } {}$
• Relative velocity is the velocity of an object as observed from a particular reference frame, and it varies dramatically with reference frame.
• Relativity is the study of how different observers measure the same phenomenon, particularly when the observers move relative to one another. Classical relativity is limited to situations where speed is less than about 1% of the speed of light (3000 km/s).