Linear Motion

Linear Motion

Momentum is defined as the product of an objects mass and the velocity with which it is moving.  It is a Vector quantity because it uses Velocity.

It is important to not mix velocity with speed, Velocity is a VECTOR which means it measures a change of distance as well as a direction where as SPEED only measures a change in distance therefore making it a scalar quantity.

Linear motion is that of movement in a straight line rather than angular motion that revolves around an axis.  There are different units and conversions for angular motion and is discussed on a different page.

Definitions
It is probably best to start this page clarifying some words that are used regularly.  If you feel you know these words well then skip this bit.  But it is always a good refresher.

Distance is the measurement in cm/m/miles etc that an object has traveled whatever the form of path it takes.  A car moving 1 mile may have done this in a straight line, curved road or even in a circle and finished where it started.  It will still have moved 1 mile. 

Displacement is the distance in a straight line between the start and end points of motion.  Thus the car moving 1 mile in a straight line its displacement will be 1 mile whereas the car going in a circle will have no displacement as it ends where it started.

Uniform meaning the motion is constant with nothing hindering the movement at any point.  A steady rate of change. 

Time a measurement that does not measure motion but the process of existence.

The Basics

It is important to inform you of how and where momentum comes from, as why it is important to look back at other topics such as Newtonian Law to relate the principle of linear motion. Momentum is taken from Newtons 3rd Law and follows the format

M=momentum

m=mass         

v=velocity      

Remember that momentum of a body remains the same until an external force acts upon it.  Meaning an object will remain at rest until it is moved.  Velocity and Acceleration can still be measured, when not in motion the object will still have motion, measured at 0.

Linear Velocity is defined as :

Linear Acceleration is defined as:

When calculating linear motion it is always taken that acceleration is constant rate. The equations form a graphical representation from which variables can be measured from. More about this soon.

Units of motion

First it should be mentioned there are specific SI units that need to be remembered and the units involved.

Velocity   ~   m/s  

Acceleration  ~  m/s/s 

s   =  distance

u  =  initial velocity

v  =  final velocity

a  =  acceleration

t  =  time

Measuring motion

The graph bellow is showing how acceleration, velocity and retardation is represented, and it is good to note that when doing calculations it is useful to draw this graph so that you can visualise the event.  

The area below the line, shaded area, represents the distance traveled over time. Slope A is the acceleration of the object, in the case on this graph it is starting from rest.  B relates to the constant velocity of the object in motion and retardation is slope 'C' this is the object coming to rest (deceleration).  

To help show how an area can relate to distance an example is given below:

The area under the graph can be divided into two triangles and one rectangle.

The area of triangle A is half base x height

            = 0·5 x 10 x 20

= 100.

The area of triangle C

                      = 0·5 x (70 - 30) x 20

= 400.

The area of rectangle B

              = (30 - 10) x 20

= 400.

The distance travelled is the total area = A + B + C

              = 100 + 400 + 400

= 900 m.

Now for the equations.

There are four equations that are used in linear motion.  These are now going to be explained, how they are derived and why.  It is important that engineers know where the equations are formed from as sometimes they are built into other equations. this next section is long but take your time reading and make sure you understand what is being taught. 

Common benchmark velocity values for measuring acceleration are:

0 - 60 mph = 0 - 26.8 m/s or 0 - 96.6 km/h

0 - 100 km/h = 0 - 27.8 m/s or 62.1 mph

Calculating average Velocity is simple. 

where:

v = velocity (m/s)

s = distance traveled (m)

t = time taken (s)

A car travels a distance of 60 km in one hour, average velocity can be calculated as

v = (60 km) (1000 m/km)  / ((1 hour) (3600 s/hour)) 

V = 16.7 m/s

Calculating Average Acceleration

Once you have the Velocity and time the journey took then it is possible to work out the average acceleration.

where

a = acceleration (m/s²)

v = change in car velocity (m/s)  

t = time of acceleration (s)

Now this only works if you are measuring the distance for which the object was increasing the speed.  If there is a section of this time where the object was at constant speed then the acceleration is incorrect.

The Linear Motion Equations

As long as the acceleration is uniform the velocity will increase as a straight line graph.  Given that  acceleration is ‘the rate of change in velocity with respect to time’ and for motions of this kind ‘u’ represents the velocity at time t=0 (the beginning) and ‘v’ represents the velocity after a time lapse t, we can formulate the following equations after rearranging to make V the subject. This form is used as 'a' is the gradient of the slope.

The distance traveled is measured by calculating the area under the graph as shown previously.

Because the equation above is reliant of the variable of velocity  being available which is not always possible, substituting the first equation in place of V will create the following equation that means distance may be calculated.

Finally to eliminate 't' from the formula combine equations 1 & 2 to create the 4th equation.

This is the set of equations that you will use for linear equations.  The angular equations are similar but use different notation.  It is important that you are able to transpose the equations as many times the values you have will not fit the original form.  There is a page on this blog that works through transposition. 

When faced with an equation for linear motion it is easy to solve if a simple method is followed.  I will now walk through an example with you so that you can learn the method which will make life simpler for you.

Example

1.    

Read through the information and create the graph so that you can see the motion that the vehicle is taking.

Now that you can see what you have you are able to seperate the questions.  For a) you need to collect together the available terms in relation to the question and allocate the correct SI unit to them.  This would be the slope rising up as you are calculating acceleration properties.

So for a)   

                       u = 0 m/s

                       v = 9 m/s

                       a = 2 m/s

Now that you have the values that you need for this question then consider the equation you would use.   There are three values available and you are looking for 'time' therefore the easiest equation to use with a transposition is 

v = u + at

If you are not confident with transposition then there will soon be a page for you to read. 

After rearranging the equation to:

t = v - u / a

Simply substitute your values in the correct places. 

t = 9 - 0 / 2

Which results in t = 4.5 seconds. 

Add this into your graph and you can see how the whole story is starting to form.

The new figure is highlighted in red to show where it belongs.  Now to move onto part b) of the equation.

The rate of retardation.  This is the same as the acceleration but this time you will be looking for 'a'.  It may look like there is a variable missing but there is not.  The question tells you that the whole motion takes 24.5 seconds.  If the first section takes 4.5 s and the second section takes 15 s its easy to calculate the last section. 

24.5 - ( 4.5 + 15) = 5

Therefore use the same equation and transpose for 'a'.

a = v - u / t

a = 1-9 / 5

a = -1.6 m/s

This is the retardation, it has a negative value as it is the slowing down of, which means there is a negative response.

Return to your graph and add the new values.

As is evident now the only information missing is the total distance traveled which is part c) of the question. 

To make this easy make it three sections and do a sum for each.  There is a, b and c.  The values for each are:

a)     u = 0                                  b)    u = 9                                 c)   u = 9

         v = 9                                         v = 9                                        v = 1

         t = 4.5                                       t = 15                                      t = 5

The equation that would be best for this process is 

s = (u + v / 2) t

a)    s = ( 0 + 9 / 2 ) 4.5                 b)  s = ( 9 + 9 / 2 ) 15                   c)    s = ( 9 + 1 / 2 ) 5

        s = ( 9 / 2 ) 4.5                             s = ( 19 / 2 ) 15                              s = ( 10 / 2 ) 5

        s = 4.5 x 4.5                                 s = 9 x 15                                       s = 5 x 5

        s = 20.25 m                                 s = 135 m                                       s = 25 m

Now you have the result of each section add them together for the total.

20.25 + 135 + 25 = 180.25 m 

So the results for the question are:

a)     4.5 seconds

b)     - 1.6 m/s

c)     180.25 m

Following is a table that will aid you in converting or understanding the difference between measurements. This will help you when working with questions to gauge how near your answers are.  This is because the calculator is not always right and common sense has to prevail when writing an answer.

Your Turn

Answer these following questions and submit.  It is beneficial for you to complete the form and test yourself rather than looking for answers.  If you would like an email to give you feedback please leave your email in the form where reuested.

Linear Motion Quiz


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