Geotechnical Engineering: Understanding the Foundation of Infrastructure

 

Kinematics

The concept of Kinematics is the study of motion without the force causing them and the mass of the body. Suppose a body of mass 2kg is pulled a distance of 10m by a force of 5N; kinematics studies only the movement in a giving direction without concern on the 5N force or the 2kg mass of the body.

The motion of a body can either be in any of these forms:

1. Random: Example is gaseous particles,

2. Translational: Example is a moving car,

3. Rotational: Example is a ceiling fan,

4. Oscillatory: Example is swinging pendulum.

Straight Line Motion (SLM)

This is the motion of a body in a straight line. There are four parameters involved in the study of this motion of a body in a straight line. They are Displacement or Distance, Velocity, Acceleration and Time.

Displacement

This is the distance travelled in a specified direction. It is the shortest distance between two points. In kinematics it is represented with (s) and is mathematically related to velocity thus:

s = vt

Where: s = displacement

v = velocity

t = time

The unit of displacement is metre (m).

Example 1

A car with a uniform velocity of 10 m/s left Enugu at 10:00 am and arrived Nsukka at 10:30 am. Calculate its displacement.

Solution:

First let’s write down the terms given to us. We have:

v = 10 m/s

T = (10:30 – 10:00) = 30 minutes, Note: we need to further change this time to be in seconds. So 30 minutes would be:

(30 x 60) sec = 1800 s.

So from the formula, we have that

s = vt

s = 10 x 1800

s = 18000 m.

Displacement and Distance

Most often students who are new to physics find it very difficult to understand the difference between displacement and distance. We will quickly highlight the differences before we continue with the next section.

Although these two quantities appear to be similar but displacement is a vector quantity and by so it is direction- aware while distance is a scalar quantity.

To fully understand the terms scalar and vector in physics, we will go ahead and define the meaning of these terms.

Scalars are quantities that have only magnitude

Vectors are quantities that have both magnitude and direction

A simple illustration of distance and displacement example is; if a man travels in a rectangular path from a starting point along the path and ends on the same spot he started, the displacement in this situation is zero while the distance is the length of the rectangular path from the starting point to the end point.

It is also important to note that when an object changes direction and start moving in opposite direction it will effectively cancel its initial displacement.

Distance can be measured by one of the following measuring tools and its unit of measurement is in meters.

Use of string

Metre rule

Vernier callipers

Micrometre screw gauge

Velocity

Velocity is speed measured in a specified direction. Many misunderstand Speed to be Velocity. In Physics, the two mean different things. Speed is the rate of change of distance with time. Velocity is a vector quantity (i.e. it has both magnitude and direction) while speed is a scalar quantity (it has magnitude but no direction) - we shall treat it in the next tutorial.

The unit of velocity is metre per second (m/s). Mathematically:

Velocity (v) = s/t

Where s = displacement, t = time.

Uniform velocity: A body is said to be in a uniform velocity when the ration (s/t) is constant - unchanging.

Average Velocity (aV): Average velocity is the total distance travelled (Stotal ) divided by the total time taken (Ttotal).

aV = Stotal /Ttotal

Example 1

Calculate the velocity of a car that covered a distance of 250 metres in 25 seconds.

Solution:

We write down the given terms as always:

S = 250m

T = 25s

v = s/t

v = 250/25

v = 10 m/s.

Acceleration

Acceleration is the rate of change of velocity with time. The unit of acceleration is metre per second squared (m/s2). Acceleration like velocity is a vector quantity that is why it is wrong to think that acceleration means the rate of change of speed with time. Mathematically Acceleration (a) is:

a = v/t

Where v = velocity, t = time.

Uniform acceleration: If the rate of change of velocity is constant, the acceleration is said to be uniform. This implies that:

a = v/t = constant.

If the velocity of a body is increasing with time, it is said to be accelerating, but if the velocity of a body is decreasing with time, it is said to be retarding or decelerating or experiencing retardation or deceleration. And then in such a case, it is important we notice the acceleration is said to be negative.

Example I

A body rolled from point X to point Y in 2 seconds with a velocity of 5m/s. Calculate the acceleration of the car.

Solution:

First, we note down our terms,

t = 2 s

v = 5 m/s

From the formula a = v/t

a = 5/2

a = 2.5 m/s2

The Equation of Motion

Deriving the equations of motion, it was noticed that if a body starts with initial velocity ‘u’ accelerates uniformly along a straight line with acceleration ‘a’ and covers a distance ‘s’ in a time ‘t’ when its velocity reaches a final value ‘v’, then the distance ‘s’ covered is given by s = average velocity times time.

s = u – v x t ……… (Equation 1)

Also by definition, acceleration ‘a’ = rate of change of velocity and since ‘a’ is constant we will have

a = (v-u)/t = v = u + at …………… (Equation II)

Eliminating t from (Equation I) and (Equation II), we fined

v2 = u2 + 2as ……… (Equation III)

Eliminating ‘v’ from (Equation I) and (Equation II), we find

s = ut + 1/2 at2 ……… ( Equation IV)

Note:

These four equations of motion are used in solving problems associated with uniformly accelerated motion. When using them, the following points should be noted.

I. Ensure that all the units match. i.e., velocity in m/s, distance in m, acceleration in m/s2 and time in s.

II. Each of the equations contains four of the five variables u, v, s, a, t. The values of three are normally given and the values of one or both unknowns are required to be found.

III. To determine which equation to use in solving a particular problem,

(a) Note down the given variables,

(b) Note down the required variable,

(c) Note down the un-given variable

(d) Then the equation to use is the one that does NOT contain the un-given variable.

Hint:

Equation I does not contain a

Equation II does not contain s

Equation III does not contain t

Equation IV does not contain v

IV. Note the conversion formulae: 1km/h = 1000/ (60 x 60) m/s OR 36km/h = 10m/s OR 1m/s = 3.6km/h.

V. Do not confuse s for distance with s the unit of time.

Example I

A car moves from rest with an acceleration of 0.2m/s2. Find its velocity when it has moved a distance of 50m.

Solution:

We note down the given variables, which are:

a = 0.2m/s²

s = 50m

The required variable is ‘v’ and

The un-given variable is t and the equation that doesn’t contain t is Equation III. So we use Equation III.

v² = u² + 2as

We should also bear in mind that the initial velocity u of a moving body which started at rest is zero.

So with that now, our u here will be 0m/s.

v² = 0² + 2 x 0.2 x 50

v = √ (0 + 2 x 0.2 x 50)

v = √ 20

v = 4.47m/s.

Example II

A car is uniformly retarded and brought to rest from a velocity of 36km/h in 5sec. Find:

(a) Its retardation

(b) The distance covered during this period.

Solution:

Like before, write down the variables. Then we have

v = 36km/h

t = 5s

s = ?

(

a) First, we should notice that by retardation, we are required to find the acceleration. Then the formula for acceleration is

a = v/t

and here we have v = 36km/h, we will need to change it to m/s before we proceed. So 36km/h = (36 x 1000) / (60 x 60) = 36000/3600 = 10m/s

Therefore we now have

a = 10/5

a= 2m/s2

(b) Now we have the acceleration, we need to find the distance covered, so we use (Equation IV). And I need someone to tell us why we should use Equation IV.

So from equation IV

s = ut + 1/2at2

Note: initial velocity (u) in this case is zero, then

Then s = 0 + 1/2 x 2 x 25

s = 25 m.

Now let’s have one or two problem set for you to solve. Please try your hand on these questions. It will help you to fully understand this topic. Once you are done with it you can jump over to the next tutorial.

Question I

A train slows from 108km/h with a uniform retardation of 5m/s2. How long will it take to reach 18km/h?

Question II

What is the distance covered in question 1?

You should try and solve the problem. Practising is a good way of learning.