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.

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