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## Sunday, January 16, 2011

### 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.

### 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 + 0.5*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.

## Sunday, January 2, 2011

### What is the voltmeter?

The voltmeter is a device that is used to measure the electromotive force (e.m.f.) or the potential difference across a circuit component.

Fig 1 below shows a voltmeter that is connected to a dry cell.

Fig 1

The voltmeter is always connected in parallel to the power supply. That is the positive terminal of the power supply is connected to the positive terminal of the voltmeter. The negative terminal of the power supply is likewise connected to the negative terminal of the voltmeter.

Please note that by convention the positive terminal is red while the negative terminal is black.

Fig 2  is the same circuit but this time using symbols.

Fig 2

Should there be another circuit element like a bulb the voltmeter is connected in parallel to the bulb, if you want to measure the potential difference across the bulb,  of across the power supply if you want to measure the potential difference across the power supply while it is supplying current to the bulb.Again with the positive terminal of the power supply connected to the positive terminal of the voltmeter and the negative terminal of the power supply connected to the negative terminal of the voltmeter. Fig 3 below illustrates the two concepts.

Fig 3

In addition to the voltmeter in Fig 1 there are also voltmeters that are dual range.  That is they can measure voltages over two different ranges. A typical high school voltmeter has the 0 – 5 V and 0 – 15 V ranges. An examples can be seen in fig 4 below.

Fig 4

The way to use the voltmeter is to connect the negative terminal of the voltmeter to the negative terminal of the power supply and positive terminal of the power supply to either the 5 V positive terminal and the 15 V positive terminal depending on which range to be used. Fig 5 shows the voltmeter being used over the 0 – 5 V range while fig 6 shows the voltmeter being used over the 0 –15 V range.

Fig 5

Fig 6

## Thursday, December 16, 2010

### Errors and uncertainties

Rules to perform mathematical operations

lg and ln

Uncertainty and errors

Multiplication and division

root, powers, etc

Errors

What is an error?

What is a zero error?

What is a parallax error?

## Wednesday, December 15, 2010

### Current electricity

What is a complete circuit?

What is an ammeter?

What is the electromotive force (e.m.f.)

What is the potential difference?

What is the kilowatt-hour?

What is the voltmeter?

### What is a capacitor?

A capacitor is simply an device that is composed of two metal conductors that are separated by an insulator. The insulator is called a dielectric. Fig 1 below shows a simplified version of a capacitor.

Fig 1

In fig 2 below two metal foils are separated by a paper dielectric. The three are them are then rolled as shown.

Fig 2

The fig 3 below shows a capacitor as sold commercially. It is a capsule that contains the rolled up aluminium foils.

Fig 3

The two plates are then connected to a power supply. The positive plate is connected to the positive terminal while the other negative plate is connected to the negative terminal as shown in fig 4 below.

Fig 5

After the capacitor is connected to the power supply, the capacitor will be charged by the power supply. That is charges will accumulate on the two plates. Positive charges will accumulate on the positive plate of the capacitor while negative charges will accumulate on the negative plate of the capacitor. The charges on the two plates will create an electric field between the two plates where the energy will be stored in the form of electric potential energy.

The fig 6 below shows a capacitor whose terminals are connected to a light bulb.

When this happens the energy stored in the capacitor will be released i.e  is “discharged” into the bulb.

Fig 6

Hence a capacitor can be simply said to be an electronic device that stores charges. It is used in rectification circuits that changes alternating current to direct current, in tuning circuits and in filter circuits to filter out direct current.

### Capacitors

What is a capacitor?