## Motion in a plane in space

An ordered pair of coordinates, e.g. (7.3m, 2.5m), determines a point in a plane and an arrow from the origin to that point is described as a **position vector** and can be used to describe the location of any point. The length of the position vector (the **magnitude**) records the distance from the origin to the point. An orientation records the angle between the position vector and some specified line (usually the *x*-axis).

The position vector of any point (x,y) can be specified as **r **= (*x, y*)

The origin is known as the **zero vector**, **0**

The longest side of any right angled triangle, known as the hypotenuse, is given by **Pythagoras’s Theorem**,

*c ^{2} = a^{2} + b^{2} *or

*c = (a*

^{2}+b^{2})^{1/2}In addition, the ratios of the lengths *a, b* and *c* define the **trigonometric ratios** provided that *θ* < 90°

sin θ = *b/c*

cos θ = *a/c*

tan θ = *b/a*

A useful result that follows from Pythagoras’s Theorem is that (sin *θ*)^{2} + (cos *θ*)^{2} = 1

This is called an *identity* since it is true for all values of *θ*, and is usually written in the form,

sin^{2} *θ* + cos^{2} *θ *= 1

Some useful values (worth remembering) are;

sin 0° = 0 | cos 0° = 1 | tan 0° = 1 |

sin 30° = ½ | cos 30° = √3/2 | tan 30° = 1/√3 |

sin 45° = 1/√2 | cos 45° = 1/√2 | tan 45° = 1 |

sin 60° = √3/2 | cos 60° = ½ | tan 60° = √3 |

sin 90° = 1 | cos 90° = 0 | tan 90° is undefined |

In the case of a position vector it follows from Pythagoras’s theorem that

*r* = *(x ^{2} + y^{2})^{1/2}*

and

*x = r *cos* θ*

*y = r *sin* θ*

θ = arctan (*y/x*)

The displacement, s_{x} = Δx = x_{2}-x_{1} and s_{y} = Δy = y_{2}-y_{1}

So the displacement vector **s** = (s_{x}, s_{y}) = Δ**r** =** r _{2}-r_{1}** = (x

_{2}-x

_{1}, y

_{2}-y

_{1})

The magnitude of the position vector is,

s = (s_{x}^{2} – s_{y}^{2})^{1/2}

And when the angle between the x-axis and the displacement is *θ*

s_{x} = *s* cos *θ*

s_{y} = *s* sin *θ*

If ** a** and

**are displacement vectors, if the arrow representing**

*b***is drawn from the head of the arrow representing**

*b***, then an arrow drawn from the tail of**

*a***to the head of**

*a***represents their resultant, the displacement**

*b***=**

*c***. This is sometimes known as the**

*a + b***triangle rule**for adding displacement vectors.

** c** =

**means that**

*a + b**c*and

_{x}= a_{x}+ b_{x}*c*

_{y}= a_{y}+ b_{y}If a displacement vector represents the change of position in a time interval, the **average velocity vector** in that time interval is

**<v>** = Δ** r** / Δ t = (Δx/Δt, Δy/Δt)

i.e.

*v **= d r/dt = (dx/dt,dy/dt)*

The speed of a particle is given by the magnitude of the vector,

*v = (v _{x}^{2} + v_{y}^{2})*

The instantaneous **acceleration vector** is the derivative of the velocity vector,

**a **= d**v**/dt = (dv_{x}/dt, dv_{y}/dt)

and the acceleration has a magnitude,

*a = (a _{x}^{2} + a_{y}^{2})^{1/2}*

and the acceleration components are,

*a _{x} – a *cos

*θ*and

*a*=

_{y}*a*sin

*θ*

The essential points that Galileo had grasped was that, the horizontal and vertical movements are independent, apart from the fact that they must both have the same direction 🙂

The vertical component of a projectile’s motion is an example of uniformly accelerated motion, in which the constant acceleration is due to the Earth’s gravity, directed downward, and has a magnitude of *g* (≈ 9.8ms^{-1}). The horizontal component is an example of uniform motion, in which the constant horizontal velocity is determined when the projectile is launched.

The time of a projectile’s flight is,

*T *= (2*u* sin*θ*)/*g*

The range is

*R* = | (2*u*^{2} sin*θ* cos*θ* )/*g*|

And for a fixed launch speed *u, *the maximum range *R _{max}* is

*R _{max} = u^{2}/g*

And is achieved when the launch angle is 45°

The path, or **trajectory**, followed by a projectile is

s_{y} = as_{x}^{2} + bs_{x}

where *a* and *b* are constants given by,

*a = -g/(2u _{x}^{2}) *and

*b = u*

_{y}/u_{x}Hopefully, you can see this form of equation is a quadratic function and therefore; A projectile travelling close to the Earth (so that *g* remains constant), and in the absence of air resistance, follows a trajectory that has the form of a parabola. A Quadratic Equation has the form

a*x ^{2} + *b

*x +*c = 0

its two solutions are always given by

*x* = *(-b± (b ^{2}-4ac)^{1/2})/2a*

Note:

if *b ^{2} > 4ac* then the formula gives two solutions

if *b ^{2} = 4ac* the two solutions are identical

if *b ^{2} < 4ac* then the root are complex (i.e. as in complex number, not as in ‘complicated’ 🙂 )

Quadratics can also be solved by **factorization**, i.e. written in the form

*a(x – α)(x – *β*) = 0*

where α + β = *-b/a *and *α*β = *c/a*

For bodies launched from a height,* *

*T = (u _{y} + (u_{y}^{2} + 2gh)^{1/2})/g*

And

*R = |(u _{x}u_{y} + u_{x}(u_{y}^{2}+2gh)^{1/2})/g|*