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## Friday, February 26, 2010

### Conservation of kinetic energy in collisions

As we have seen in this post during a collision between two object the velocity of the two colliding objects changes. Hence you would agree that he kinetic energy of the objects would change.

We have also seen in this post that in all collisions the sum of linear momentum is a constant. This is the principle of conservation of linear momentum.  However as we are going to see in some collisions, elastic collisions, the sum of kinetic energy is also constant. That is the sum of the kinetic energies of the objects before the collision is the same as the sum of the kinetic energies after the collision.

We are now going to see an example of how to use this “principle” which i am going to call the “principle of conservation of kinetic energy”.

Fig 1 below shows two objects travelling towards each other and fig 2 shows the two objects separating after the two objects have separated.

So if the principle of conservation of kinetic energy applies then it means that Sum of kinetic energy before collision = sum of kinetic energy after collision

Fig 1 Before collision

Before the collision

Sum of kinetic energy before collision = kinetic energy of object 1 + kinetic energy of object 2

= 1/2m1u12 + 1/2m2v12

Fig 2 After collision

After the collision

Sum of kinetic energy after collision = kinetic energy of object 1 + kinetic energy of object 2

= 1/2m1u22 + 1/2m2v22

If you refer back to the principle then

Sum of kinetic energy before collision = sum of kinetic energy after collision

1/2m1u12 + 1/2m2v12   =   1/2m1u22 + 1/2m2v22

Example

Two balls of mass 1.0 kg are travelling towards each other at a speed of 5.0 m/s towards the right (ball 1)  and 5.0 m/s towards the left (ball 2) respectively. If after separation one of them moves moves off a speed of 5 m/s towards the right (ball 2 ) calculate the speed of the other ball.

Sum of kinetic energy before collision = sum of kinetic energy after collision

1/2m1u12 + 1/2m2v12   =   1/2m1u22 + 1/2m2v22

1/2*1.0*5.02 + 1/2*1.0*(-5.0)2   =   1/2*1.0*u22 + 1/2*1.0*5.02

Solving for u2 the velocity of object 1 after the collision

12.5+ 12.5 = 0.5*u22+12.5

12.5= 0.5*u22

12.5 /0.5 = 25 =u22

u22 = 25

There can be two solution to u2

First u2=-5 m/s and secondly u2= 5 m/s

Since the second ball is already travelling towards the right and according to the principle of conservation of linear momentum then the only solution is

u2=-5 m/s

## Tuesday, February 23, 2010

### Conservation of linear momentum in collisions

As we have seen in this previous post, in a close system the sum of linear momentum is a constant. Hence in any collision if the sum of linear momentum of the objects before the collision would be equal to the sum of linear momentum after the collision.

Fig 1 below shows two spherical objects travelling toward each other. The mass of the two objects are 1.0 kg each. Object 1 is travelling at a velocity of 5.0 m/s  and object 2 is travelling at a velocity of –5.0 m/s.

Fig 1

If  p1b is the linear momentum of object 1 before the collision and p2b

is the linear momentum of  object 2 before the collision then

the sum of linear momentum = p1b +p2b

= m1u1 + m2v1

=1.0*5.0 + 1.0*(-5.0)

= 5.0 - 5.0

=0 Kg m/s

As you can see before the collision the sum of the linear momentum is 0 kg m/s. According to the principle of conservation of linear momentum the sum of linear momentum at all time will thus be 0 kg m/s before and after the collision.

Fig 2 below shows the spherical objects after the collision while they are separating. According to the principle of conservation of linear momentum after the collision the sum of linear will thus be 0 kg m/s.

Question

If after the collision the two objects separate such that object 2 moves with a velocity of  5.0 m/s, calculate the velocity of object 1.

Fig 2

If  p1a is the linear momentum of object 1 after the collision and p2a

is the linear momentum of  object 2 after the collision then

the sum of linear momentum = p1a +p2a

= m1u2 + m2v2

=1.0*u2  + 1.0 *5.0

= u2  +5.0

Now if we use the principle of  of conservation of linear momentum

sum of linear momentum before the collision = sum of linear momentum before the collision

0 =   u2  +5.0

u2  = - 5.0 m/s

## Sunday, February 14, 2010

### Newton’s second law of motion

We have seen in a previous post the Newton’s first law of motion.

Today we are going to see the second law of motion.

From our discussion in the first law of motion and in the post on acceleration when a force acts on an object either velocity increases due to a change in speed or the velocity changes due to a change in direction. Remember velocity is a vector quantity hence it will change if either the direction or the magnitude changes.

Hence we can deduce that if a force acts on an object then the velocity of the object changes.The more force is applied the greater the change in velocity. This must mean that if the mass remain constant then the momentum of the object changes. And hence the more force is applied the more the linear momentum will change.

Let us look at the situation below:

A force F acts on an object of mass m for a time t such that before the application of the force the object has a velocity v1 hence a linear momentum of p1 and after the force is applied the velocity of the object is v2 hence the linear momentum is p2.

Before the force is applied velocity = v1 and linear momentum = p1

After the force is applied velocity =va2  and linear momentum is p2.

Now the Newton’s second law of motion states that the force applied is directly proportional to the rate of change of linear momentum of the object.

If = a then

If k = 1

Then this will reduce the equation to

F= ma

This where the famous equation F = ma come from. This why it is also considered as the Newton’s second law of motion

### Newton’s first law of motion

In order for an object to accelerate it needs to be acted upon by a force. As we have seen in an earlier post on acceleration.

From the post on acceleration we can deduce that there are four changes that can take place if a force acts on an objects:

1. The speed of the object will increase from zero if it was at rest and the object will move in the direction in which the force is acting.
2. The speed of the object will increase if the force is acting in the same direction as the the direction of motion of the object until the force no longer acts on the object.
3. The speed of the object will decrease if the force is acting in a direction opposite to the direction of motion of the object. The speed of the object might decrease to zero if the force acts for long enough.
4. If the object is moving and the force is acting in a direction perpendicular to the direction of motion of the object then the object will move in a circular path. When the application of the force stops the object will leave the circular path and would move in a straight line in a direction tangential to the circular path.

These four changes can be combined together to produce the Newton’s first law of motion as shown below:

Unless a force is applied an object will continue in its state of rest or uniform motion in a straight line.

Another way to write this law:

If a force is applied (i) an objects that was at rest will no longer be at rest or (ii)if moving with a uniform  will no longer have a uniform or constant speed or lastly (iii) if moving in a straight line will no longer move in  the same direction

This law is thus explaining that if a force is acting on an object then the object will experience an acceleration. However we are officially learn this only when we see the Newton’s second law of motion.

## Saturday, February 13, 2010

### What is a collisions?

A collision is a phenomenon that can be described in three distinct steps.

1. Firstly the two objects that would collide must approach each other. The two objects may be travelling towards each other as shown below in fig 1. They may also be moving in the same direction as in Fig 2 but if the first one is slower than the second one they would meet at a certain time.

Fig 1 Two objects moving towards each other directly

Fig 2 Two objects moving in the same direction but about moving closer to each other.

2. The two objects that are travelling towards each other must meet. That is they make contact. During this contact they interact by exerting forces on each other. The forces that they exerts on each other must be equal to each other according to Newton’s third law of motion.

Fig 3 The two objects make contact and exert forces on each other

3. The two objects then separate and move away from each other in opposite as shown in fig 4 or moves in the same direction but  with increasing distance of separation as shown in fig 5.

Fig 4 The two objects move away from each other

Fig 5 Two objects move in the same direction but the distance between them increases with time

### Homogeneity of equations

As we have seen in the chapter on derived units some quantities have units that is composed of two or more base units.

The physics students will meet with different equations. Now some of these equations are correct while some of them are not.

The test to check whether an equation is correct is through the homogeneity test as shown in fig 1 below.

Fig 1

So the question is how to check whether an equation is homogenous?

Example 1

Check whether the following equation is homogenous?

P =hdg

Where p is pressure. h is height, d is density and g is the acceleration due to gravity.

As you can see the equation has two sides. The left-hand side and the right hand side.

Now you  have to check that both side of the equation has the same unit.

Pressure = Force / Surface area

Unit of pressure = Unit of force/ unit of area

= kg m s-2 / m2

= kg m-1 s-2

unit of hdg = unit of h* unit of d* unit of g

= m * kg m-3 * m s-2

= kg m-1 s-2

Since both side of the equation has the same unit then the equation is homogenous.

Please note that if their is a dimensionless constant in the equation like k, 1/2 or something like that then the dimensionless constant, by definition, has no unit and as a result will not be involved when the test for homogeneity is performed.

Example 2

Check whether the following equation is homogenous.

v2 = u2 + 2as

In this case the equation is composed of three groups of quantities: v2 , u2 and 2as. You should check that the three groups of quantities have the same unit.

If the three group have the same unit the equation is homogenous otherwise it is not homogenous.

Note: Please see chan’s comment below for a more detailed  answer to example 2

## Thursday, February 11, 2010

### What are atoms?

Atoms are the basic constituent of matter. Hence matter is made of atoms arranged in a particular way . Of course there are different types of atoms and the type of atoms and the type of arrangement will determine the type of matter.

We are now going to see a simple model of the atom and what are the different constituents of the atom.

Constituents

The atom is composed of three basic constituents.

1. The proton
2. The neutron
3. The electron

All the different types of atoms will have these three constituent particles. However the number of each will differ from one type of atom to another. We can thus deduce that it is the combination of these three constituent particles that determine the nature of the atom.

However in neutral atoms the number of proton is the same as the number of electron. If the two differs it means that the atom is charged. If there are more protons that electrons then the atom is positively charged and if the number of electron is greater than the number of proton then it means that the atom is negatively charged.

The table below lists the properties of the different constituent particles of the atom.

 Property Proton Neutron Electron Mass 1.67 x 10–27 kg 1.67 x 10–27 kg 9.10 x 10-31 kg The proton and the neutron is on average 10000 times more massive that the electron charge +1.60 x1019 C No charge This is because as you will see later on in radioactivity it is composed of one proton and an electron. The two opposite charges thus cancel each other -1.60 x10-19 C Location In the nucleus In the nucleus In orbits around the nucleus

As we have seen in the table above the neutron and the proton are present in the nucleus while the electron is present in orbits around the nucleus (They are in reality different  but we assume that they are planet-like in this model).

A model of the atom is shown below in fig 1

Fig 1

This model is a mere representation of the atom. You will learn later on how the electrons would be arranged if there are more than four. The size of the nucleus is also quite small. So small that its diameter is 10000 times smaller than the radius of the atom. Thus to represent it truly on fig 1 a tiny dot would have been more appropriate.

### Linear momentum and collisions

Linear momentum

What is linear momentum?

Principle of conservation of linear momentum

What is a collision?

### Circular motion

Introduction to circular motion

What is angular velocity?

## Wednesday, February 10, 2010

### Principle of conservation of linear momentum

The linear momentum is the product of the mass of an object and the velocity of an object.

Now if you have two objects that are moving toward each other or one object explodes into two different objects at all time the sum of the linear momentum of the objects will give a constant number.

Hence the principle of conservation of linear momentum states that in a close system the sum of linear momentum is constant.

The equation will be that at all time

p1 +p2+p3+p4+p5+ ………= constant

Consider the diagram below:

Fig 1 Before firing ball

Fig 2 After firing ball

As you can see from the diagram above, you have a close system that is made up of the the cannon and the cannon ball. Hence at all time the sum of the linear momentum will be a constant. Before the firing of the ball and after the firing of the ball.

Example

A canon ball is fired such that the ball leaves the cannon at a velocity of  150 m/s. If the cannon has a mass of 1000 kg and the cannon ball has a mass of 15 kg, determine the velocity of recoil of the cannon.

Hence if you apply the principle of conservation of linear momentum.

Before firing cannon ball

Sum of linear momentum before firing = pic  +pib

where pic is the initial momentum of the cannon and pib is the initial momentum of the ball.

After firing the cannon ball

sum of linear momentum after firing= pfc  +pfb

where pfc is the final momentum of the cannon and pfb is the final momentum of the cannon ball.

We are now going to determine the speed of recoil of the cannon after recoil.

Sum of linear momentum before firing = pic  +pib

= 1000*0 + 15*0

= 0 kg m/s

sum of linear momentum after firing= pfc  +pfb

=1000*vrc +15*150

=1000*vrc +2250

where is the vrc velocity or recoil of the cannon.

Hence if the sum of linear momentum is a constant

then the sum of linear momentum before firing = sum of linear momentum after firing

0 = 1000*vrc +2250

0 – 2250 = 1000*vrc

-2250 = 1000*vrc

vrc = –2.25 m/s

This principles is used also when collisions are considered. We shall see collisions in a future post.

### Prefixes in Physics

You have seen in a previous post that physical quantities are properties of an object that can be measured with a measuring instrument.

Hence when you take the measurement the number that is obtained might be unusually large or small so that it is difficult to work with them. We have already seen such an example with the electron volt.

Hence in order to make working with these numbers we use prefixes. A prefix is a symbol that you place in front of a unit. The prefix will indicate a multiple of the unit.

The table below Let us have a look at a list of prefixes that you may encounter in your physics classes.

 Name of prefix Symbol Multiple of Giga G x109 Mega M x106 kilo k x103 Hecto H x102 deci d x10-2 milli m x10-3 micro Î¼ x10-6 nano n x10-9

So how to use a prefix?

Example 1

Suppose a measurement is made and the result is as shown below.

Power = 2000000 W

Now this power can be rewritten in the form of standard form as shown below:

Power = 2.0 x 106 W

As we have seen in the table the prefix Mega is x106 meaning 1 million.

Hence

Power = 2.0 x 106 W

can be rewritten as

Power = 2.0 (x 106) W

=2.0 MW

If 1 mega is one million then 2.0MW is 2 million W.

Example 2

The diameter of a cell is 0.000002 m

If you rewrite this in standard form then the diameter will be written like this

diameter = 2.0 x10-6 m

if you would recall the prefix micro is represented by x10-6 meaning 1 millionth.

Hence

diameter = 2.0 x10-6 m

can be rewritten as

diameter = 2.0 x10-6 m

diameter = 2.0( x10-6 )m

=2.0  Î¼m

If a micro is one millionth then 2.0  Î¼m is 2 millionth of a meter.

However it is also possible that a measurement is taken but it is impossible to write it directly in term of a prefix.

The wavelength of red light= 6.50 x10-7 m

As you can see it is not possible to write it directly in term of either micro or nano. In this case a division must be performed. If you want to write the wavelength of red light in term of micro then you have to divide the wavelength by 1 x10-6 but if you want to write it in term of nano you will then have to divide it by 1 x10-9.

To write the wavelength in term of micro

wavelength = 6.50 x10-7 m

= 6.50 x10-7/ 1 x10-6 Î¼m

=0.650 Î¼m

To write the wavelength in term of nano

wavelength = 6.50 x10-7 m

= 6.50 x10-7/ 1 x10-9 nm

=650 Î¼m

As you can see it is quite easy to write any number in term of prefix. If after this post you still have difficulties, then you can leave your question in the comment section below.

## Sunday, February 7, 2010

### How to calculate the energy of a photon?

You would remember from a previous post that a photon is a packet of energy. Hence all electromagnetic radiations are emitted in term of photons.

We also saw that the energy of a photon is given by the equation

E = hf

where h is the Planck’s constant and f is the frequency of the electromagnetic radiation’s photon.

Fig 1

From the equation we can thus deduce that a photon’s energy is directly proportional to its frequency.

Hence the higher the frequency of a photon the greater the energy it has.

From fig 1 we can deduce that since radio wave has the least frequency then the photons of radio waves has the least energy. And as shown in fig 1 as we move from left to right across the electromagnetic spectrum the frequency increases. This means that the energy of the photon also increases. Photons of gamma rays would thus be the most energetic.

Example

Calculate the energy of a photon of red light of wavelength 650 nm if the speed of light is 3.0 x 108 m/s and Planck’s constant = 06.63 x 10-34 m2 kg s-1.

Now if you remember the frequency f, the wavelength Î» and the speed  c of all electromagnetic radiations are related by the equation

c = fÎ»

hence making f the subject of formula

f =c/Î»

Calculating f

f = 3.0 x 108 /650 x 10-9

= 4.6 x 1014 Hz

Calculating the energy of the photon using the equation E = hf

Energy E = 6.63 x 10-34 * 4.6 x 1014

=3.0 x10-19 J

You would remember in a previous post that we discussed about the use of the electron volt eV as unit of energy in quantum mechanics.

We are now going to see how to determine the energy of the photon in eV

E = 3.0 x10-19 J

= 3.0 x10-19 /1.6 x10-19

= 1.9 eV

As we have seen in the previous post working with eV is much better.

### Third equation of motion

We have seen in two earlier post the  first equation of motion and the second equation of motion. If you have not read these two post I would suggest you do so before continuing.

The third equation is derived from the first and the second equations. The mathematics is straight forward as I will show below.

To derive the third law we are going to use the first equation of motion as shown below

v = u +at                             Equation 1

and this second equation that we used when we derived the second equation. If you have forget check the post on the second equation of motion.

s = 1/2( v + u)*t                      Equation 2

Now if we change the first equation of motion we will have

Equation 3

We can then substitute equation 3 above in equation 2 to obtain

Multiplying on both side by 2a

Simplifying further

Removing brackets

Simplifying further

So this is the derivation for the third equation of motion.

Example A car travelling at a speed of 10 m/s brakes with a deceleration of 2 m/s2 to a stop. Calculate the distance travelled by the car during the deceleration.

To answer this question we first have to identify the different variables.

distance travelled s = ???

initial speed u = 10 m/s

final speed v = o m/s

acceleration = - 2 m/s2

Note that the acceleration is negative since deceleration involves a decrease in speed hence the number is positive.

We now need to use the third equation of motion as shown below.

Substituting the different values in the equation

2*-2*s = o2 - 102

-4s = –100

s =25 m

Hence the distance travelled by the object during the deceleration is 25 m

## Saturday, February 6, 2010

### What is the electron volt (eV)?

The electron volt is a unit of energy like the joule (J).

However in quantum mechanics if the joule is used, then very small numbers would be obtained. These small numbers are very clumsy to use as a result a better unit to use is the electron volt.

Fig 1

As you can see in fig 1, we have two metal plates that are connected to a power supply of e.m.f. 1 V. As a result a potential difference of 1 V will be set up be set up between the plates. An electric field will be set up between the plates as a result of the potential difference.

Now if a stationary electron is released from the negatively charged plate it will experience an acceleration due to the electric field and as a result will move towards the positively charged plate with increasing velocity.

When it would reached the positively charged plate it would have a certain velocity.

The Kinetic energy gained by the electron can be calculated by the following equation:

Kinetic energy gained Ek = Charge on particle * Potential difference between plates.

Ek = Q * V

Hence it can be deduce that if the charge is an electron then the kinetic energy that is gained by the electron is

Kinetic energy gained Ek = Charge on electron (elementary charge) * 1 V

= 1 eV

Hence the eV is the energy that is gained by an electron or any particle of charge -e or +e when accelerated by a potential difference of 1 V

If you want to know how many joule there is in 1 eV then it is as shown below

Kinetic energy gained Ek = Charge on electron (elementary charge) * 1 V

= 1.6 x 10-19* 1

= 1.6 x10-19 J

We can thus say that 1 eV is equivalent to 1.6 x 10 –19 J

As we have said above in quantum mechanics it is better to use the the eV because if the joule is used we would be dealing with small numbers which would make the works and calculations difficult.

### Second equation of motion

We have seen in a previous post the first equation of motion.

Today we are going to see the second equation of motion.

You would remember that the first equation dealt with acceleration, time taken, initial velocity and final velocity.

So how do we derive the second equation of motion?

Let us take an object that starts from an initial velocity u and then accelerates during a time t until it reaches a final velocity v.

Now during the time the object was accelerating the object has moved a distance s. In the second equation of motion we will try to derive the equation to calculate this distance s.

Fig 2 below shows what is happening to the object.

Fig 1

Now you would remember that the area under a velocity-time graph is the distance travelled by the object. Hence we are going to calculate the area under the graph as indicated by the shaded area below.

Fig 2

Now if s the distance travelled is the area under a velocity-time graph then we can say that

Area under graph = 1/2( v + u)*t

hence s = 1/2( v + u)*t

Now if you can remember the first equation of motion is

v = u +at

Hence if we replace v = u +at into s = 1/2( v + u)*t we get

s = 1/2( v + u)*t

s = 1/2( [u +at] + u)*t

s = 1/2( u + at + u)*t

s = 1/2(2 u + at )*t

s =1/2(2ut +at2)

S = ut + 1/2at2

Hence as you can see above the second equation of motion is

S = ut + 1/2at2

We are now going to look at two examples where the second equation of motion can be used.

Example 1

A bus starts from rest and accelerates at a rate of 2.5 m/s2 for a time of 50 s. Determine the distance travelled by the bus during the acceleration phase.

Now let us identify the different variables involved.

u the initial speed = 0 m/s

t time taken = 50 s

a acceleration = 2.5 m/s2

s distance travelled = ???

If we want calculate s the distance travelled we will have to use the second equation of motion

S = ut + 1/2at2

Substituting the different values we get

S = ut + 1/2at2

= o*50 + 0.5*2.5 *502

= 0 + 1.25*2500

=3125 m

Example 2

An aero plane travelling at a speed of 300 m/s lands on a track 2000 m long and decelerates for a time of 25 s until it comes to rest. Calculate the deceleration needed to stop the plane.

We must now identify the different variables involved as in the first example.

initial speed u = 300 m/s

time taken t = 25 s

Distance travelled s = 2000 m

acceleration a = ????

If we want calculate the acceleration a we will have to use the second equation of motion

S = ut + 1/2at2

and make a the subject of formula.

s – ut = 1/2at2

a =2(s - ut)/t2

= 2(2000 – 300*25)/252

= –8.8 m/s2

As you can see using the second equation of motion is not that difficult. Should you enter into any difficulties just post the question into the comment section.

## Wednesday, February 3, 2010

### Subject of formula : The basics

Subject of formula is a very important concept in physics since it will enable you to manipulate equations properly.

If you have the equation

s = ut + 0.5 at2

and you want to calculate the  acceleration a, how will you do it?

You will have to write an equation of a in term of the other quantities like

a =

You will do this by moving the different quantities around until only a is left on one side. This is called making a the subject of formula.

We will start by studying four simple rules and then I will teach you how to use them to manipulate simple equations.

Rule 1

If you have the equation A + B = C and you want to have A in term of B and C what will you do?

A + B = C

You want A to remain on  the left-hand side, as a result you will want to move + B to the right-hand side.

The + B will have to change to – B when transferred to the right-hand side.

Hence A + B  = C

becomes

A = C - B

The first rule is that a + on one side becomes a + on the other side.

Rule 2

If you have A – B = C and you want to have A in term of B and C what will you do ?

A – B = C

You want A to remain on the left-hand side and move – B to the right-hand side.

The – B will have to change to + B when transferred to the right-hand side.

Hence  A – B = C

becomes

A = C +B

The second rule is that a – on one side becomes a + on the other side

Rule 3

If you have A*B = C and you want to write A in term of B and C what would you do?

A*B = C

If you want A to remain on the left-hand side then you would have to move *B to the other side.

The *B will have to change to /B when transferred to the right-hand side.

Hence A*B = C

becomes

A = C/B

The third rule is that a * on one side becomes a / on the other side.

Rule 4

If you have A/B = C and  you want to write A in term of B and C what would you do?

A/B = C

If you want A to remain on the left-hand side you will have to move /B to the other side.

The /B will have to change to *B when transferred to the other side.

Hence A/B =C

becomes

A = C*B

The fourth rule is that a / on one side becomes a * on the other side.

### First equation of motion

The first equation of motion is mainly derived from the definition of acceleration.

You would remember that acceleration is the rate of change of velocity with time  and that the equation to calculate it is

acceleration = (change in velocity)/time taken

= (final velocity – initial velocity) /time taken

a  = (vf – vi)/t

Rearranging the equation will give you

vf = vi + a*t

In some books the equation is given as such

v = u + at

If we used the equation above then the symbol v will be the final velocity and the symbol u will be the initial velocity of the object.

Now as we have seen in this post on acceleration if the final velocity is greater than the initial velocity then the acceleration is positive but if the initial velocity is greater than the initial velocity then the acceleration is negative.

Now let us have a look at two examples on the first equation of motion.

Example 1

A boy starts from rest accelerate to a velocity of 10 m/s in a time of 10 s. Determine the acceleration of the boy.

His final velocity v = 10 m/s

His initial velocity u = 0 m/s since he started from rest.

Time taken t = 10 s

Using the equation v = u +at and making  a the subject of formula will give you a = (v-u)/t

Substituting the different values of v, u and t in the equation will give you

a = (v-u )/t

=(10 – 0 ) /10

= 1 m/s2

Example 2

A train travelling at a speed of 100 m/s is decelerated to a speed of 50 m/s in a time of 15 s. Calculate the deceleration that the train is subjected.

final velocity v = 50 m/s

initial velocity u = 100 m/s

time taken t = 15 s

using the equation v = u +at and making a the subject of formula will give you the equation

a = (v-u) /t

Substituting in the equation above the different values of v, u and t

a = (v-u)/t

= (50 –100 )/15

= –50 /15

= - 3.33 m/s2

Since the acceleration is – 3.33 m/s2 then

deceleration  = 3.33 m/s2

## Monday, February 1, 2010

### what is acceleration?

Acceleration is a term that is associated with a change with velocity of an object.

It is defined as the rate of change of velocity with time.

So it means that in order for an object to accelerate you need to have a change of velocity over a period of time.

There are three ways in which an object can accelerate that is its velocity can change over time.

1. Firstly as shown in fig 1 below the object’s velocity can increase with time.

Fig 1

2. Secondly as shown in fig 2 below the velocity of the object can decrease with time.

Fig 2

3. Lastly the velocity of the object can change due to a change in direction as shown below in fig 3. Remember the velocity is a vector quantity and as a result has a direction component and a magnitude component. Even if the magnitude remain constant a change in direction will result in a change in the velocity.

Fig 3

However we will only consider the first two cases today. The third one will be considered when we study circular motion.

If an object has a velocity v1 is subjected to a force such that after a period of time t of being subjected to the force its velocity becomes v2 then we can say that the object is experiencing an acceleration.

This acceleration can thus be calculated using the equation

From this equation we can thus see that when the velocity of an object increases then the acceleration is a positive number and when the velocity of the object decreases the acceleration is a negative number.

The acceleration is a vector quantity since it is a vector quantity divide by a scalar quantity.