# 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[/itex]
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

1. A ball is rolled 15m in 10s. How fast was it travelling?
2. A Car accelerates from rest to 25ms$^{-1}$ while travelling 15m. How long did this take?
3. An athlete accelerates from rest to 10ms$^{-1}$ in 1s. State the acceleration.
4. Calculate the acceleration created when a 10kg block has a force of 200N exerted on it.
5. Using the acceleration found above, how fast will a object travelling at 10ms$^{-1}$ be travelling after 10 seconds?
6. 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.
7. 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.

1. 1.5ms$^{-1}$
2. 1.2s
3. 10ms$^{-2}$
4. 20ms$^{-2}$
5. 210ms$^{-1}$
6. $\underline{s}=13\underline{i}-403.1\underline{j}-13\underline{k}$. And therefore the distance is 403.52m (2dp)
7. $v=(2t-16)ms^{-1}$ 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.