# Introduction to mechanics
## Introduction

Mechanics is the study of the motion of objects using equations. This can be achieved by using standard formulae or mathematical techniques such as differentiation or integration.

## Definitions

A few definitions need to be discussed.
* $Displacement$ - How far an object is from its starting position. This is given the symbol $s$
* $Distance$ - How far an object has travelled. This is not usually used in calculations.
* $Initial$ $Velocity$ - The starting rate of change of displacement. This is given the symbol $u$
* $Final$ $Velocity$ - The ending rate of change of displacement. This is given the symbol $v$
* $Speed$ - The rate of change of distance. This is not usually used in calculations.*
* $Acceleration$ - The rate of change of velocity. This is given the symbol $a$
* $Time$ - How long has elapsed from the start of the event. This is given the symbol $t$
Speed can be calculated as the absolute value of velocity, ie spd = |u| or spd = |v|

## Standard Units

The SI units (standard units) of the quantities used in this section are given below:
* $Displacement$ - metres $(m)$
* $Velocity$ - metres per second $(ms^{-1})$
* $Acceleration$ - metres per second squared$(ms^{-2})$
* $Time$ - seconds $(s)$

## Motion in a straight line with constant velocity

To calculate the displacement moved by an object with constant velocity we use the following formula:
$s=ut$
Example: A car is travelling at 10ms$^{-1}$. How far will it travel in 12s? s = 10 x 12 = 120. Thus the car will travel 120m in 12s.
Note. This formula is based on a formula shown below.

## Newton's equations of motion

In mechanics, the following formulae are VERY important. These are:
* $v=u+at$
* $s=ut+0.5at^{2}$
* $s=0.5t(u+v)$
* $v^{2}=u^{2}+2as$
These can be used to solve any motion question where there is a constant (or assumed to be constant) acceleration. Even if the acceleration is 0. The equation from the above section $(s=ut)$ is derived from $s=ut+0.5at^{2}$ with $a=0$ and therefore $s=ut+0$ etc. To decide which equation to use, look at the data you are given. Write down a list of the quantities given (the s,u,v,a and t values). Then cross out the quantity that is not mentioned in the question. All you have to do now is see which equation has the 4 quantities that you have left in and then solve.
Example 1: A car is travelling at 10ms$^{-1}$ when it starts accelerating at 1.5ms$^{-2}$. How far will it travel in 15s? s = ?, u = 10, v =, a = 1.5, t = 15
Thus use $s=ut+0.5at^{2}$. s = (10 x 15) + (0.5 x 1.5 x 15²) s = 150 + 168.75 = 318.75 Thus the car will travel 318.75m
Example 2: A cyclist is initially at rest and then accelerates to 12ms$^{-1}$ in 6s. Calculate the acceleration. s =, u = 0, v = 12, a = ?, t = 6
Thus use $v=u+at$. 12 = 0 + (a x 6) 12 = 6a Thus a = 2 The cyclist is accelerating at 2ms$^{-2}$
Example 3: A van accelerates at 1.5ms$^{-2}$ over a distance of 10m to 20ms$^{-1}$. How fast was the van travelling before it started accelerating? s = 10, u = ?, v = 20, a = 1.5, t =
Thus use $v^{2}=u^{2}+2as$. 20² = u² + (2 x 1.5 x 10) 400 = u² + 30 u² = 370 u = ${\sqrt {370}}$ = 19.2 (3sf) The van was travelling at 19.2ms$^{-1}$ (3sf)
Example 4: A lorry can come to a stop from 10ms$^{-1}$ in 3 seconds. How far will it travel before it stops? s = ?, u = 10, v = 0, a =, t = 3
Thus use $s=0.5t(u+v)$. s = (0.5 x 3) x (10 + 0) s = 1.5 x 10 s = 15 The lorry will cover 15m before stopping.

## Applied force

Whenever a force is applied to an object it accelerates proportionally to the size of the force applied.
ie $F\alpha \ a$. and therefore $F=ka$ where k is a constant.
Via calculations the constant k, is actually the mass (m) of the object being accelerated therefore: $F=ma$.
* Force is measured in Newtons (N) and is given the symbol F
* Mass is measured in Kilograms (kg) and is given the symbol m
Example: A freighter weighing 100000kg is accelerating at 10ms$^{-2}$. What force is required? Use $F=ma$ F = 100000 x 10 F = 1000000N A force of 1000kN is required. (Where 1kN = 1000N)

## Vector motion

So far we have only really looked at motion in 1 dimension. Some times you need to look at motion in 2 and maybe 3 dimensions. To achieve this, all you have to do is convert the equations of motion into vector equations. Therefore the equations of motion become:
* ${\underline {v}}={\underline {u}}+{\underline {a}}t$
* ${\underline {s}}={\underline {u}}t+0.5{\underline {a}}t^{2}$
* ${\underline {s}}=0.5t({\underline {u}}+{\underline {v}})$
Where the underlined quantities are vector quantities none-underlined quantities are scalar.
Note: the last equation of motion is missing as this requires a more complex method of solving as it involves vector products.
Example: A ball is moving with an initial velocity of $({\underline {i}}+3{\underline {j}}-6{\underline {k}})$ms$^{-1}$ for 2 seconds. Assume the only acceleration is that of gravity, which can be taken to be 9.8ms$^{-2}$. Find the position vector of the ball at time t = 2s. (i, j and k represent the x,y and z axis of the cartesean system)
${\underline {s}}$ = ? ${\underline {u}}=({\underline {i}}+3{\underline {j}}-6{\underline {k}})$ v =
${\underline {a}}=-9.8{\underline {j}}$ $t=2$ Thus use ${\underline {s}}={\underline {u}}t+0.5{\underline {a}}t^{2}$ ${\underline {s}}=2{\underline {i}}+6{\underline {j}}-12{\underline {k}})-39.2{\underline {j}}$ ${\underline {s}}=2{\underline {i}}-33.2{\underline {j}}-12{\underline {k}}$ Thus the ball is at $(2{\underline {i}}-33.2{\underline {j}}-12{\underline {k}})$
To find the distance of the position vector, take the modulus (|a|) of the vector using Pythagoras' theorem.

## Mechanics using calculus

More often than not you will be presented with a problem that gives you the displacement/velocity/acceleration as a function rather than just values. To solve these you must use differentiation and integration techniques.
Acceleration, velocity and Displacement are related as follows:
* $Displacement=\int (Velocity)dt$
* $Velocity=\int (Acceleration)dt$
And therefore..
* $Acceleration={\frac {d}{dt}}(Velocity)$
* $Velocity={\frac {d}{dt}}(Displacement)$
Example: A car's acceleration describes the function $a={\frac {4t^{3}}{3}}-{\frac {3t^{2}}{2}}+2$. State the function of the velocity given that the car was at rest before accelerating. $a={\frac {dv}{dt}}={\frac {4t^{3}}{3}}-{\frac {3t^{2}}{2}}+2$ Therefore $v=\int ({\frac {4t^{3}}{3}}-{\frac {3t^{2}}{2}}+2)dt$ $v={\frac {t^{4}}{3}}-{\frac {t^{3}}{2}}+2t+C$ where C is the constant of integration</math> At time t = 0, v = 0. Therefore: $0=0-0+0+C\Rightarrow \ C=0$ $v={\frac {t^{4}}{3}}-{\frac {t^{3}}{2}}+2t$

## Questions on introduction to mechanics
### Questions

* A ball is rolled 15m in 10s. How fast was it travelling?
* A Car accelerates from rest to 25ms$^{-1}$ while travelling 15m. How long did this take?
* An athlete accelerates from rest to 10ms$^{-1}$ in 1s. State the acceleration.
* Calculate the acceleration created when a 10kg block has a force of 200N exerted on it.
* Using the acceleration found above, how fast will a object travelling at 10ms$^{-1}$ be travelling after 10 seconds?
* A rock is thrown with velocity ${\underline {i}}-5{\underline {j}}-{\underline {k}}$. Assuming that the only acceleration acting upon the rock is gravity, (9.8ms$^{-2}$) state the position vector after 13s and therefore the distance after 13s.
* An object's path is described by the function $t^{2}-16t+64$. Find the velocity function and therefore the acceleration applied to the object.

### Answers

* 1.5ms$^{-1}$
* 1.2s
* 10ms$^{-2}$
* 20ms$^{-2}$
* 210ms$^{-1}$
* ${\underline {s}}=13{\underline {i}}-403.1{\underline {j}}-13{\underline {k}}$. And therefore the distance is 403.52m (2dp)
* $v=(2t-16)ms^{-1}$ and therefore $a=2ms^{-2}$.

## Credits

Written by Richardtebbs 00:21, 23 January 2007 (UTC) Please find any mistakes and update them to keep content accurate.