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.

Types of distance-time graph

As we have seen in this post on distance-time graph, it is easier to extract information from a distance-time graph than from a paragraph or from a table of time and distance travelled.

We have seen in this post the gradient of the distance-time graph is the speed. Hence you must be able to deduce how the speed of an object varies according to the shape of its distance time graph.

Let us now look at some distance-time graph and see how the speed varies with time. We will then look at the corresponding speed-time graph for the object.

1. Fig 1 below shows a distance-time graph shows that the distance travelled by the object from time  0 s to t s  is 0 m. This means that the object has not moved at all and as a result it is motionless at the starting point.

The gradient of the graph is 0, hence it means that form time o s to time t s, the speed is 0 m/s.

The speed-time graph is thus as shown in fig 2.

Fig 1

Fig 2

2. Fig 3 below is a distance-time graph that shows an object whose distance travelled is constant form 0 s to time t s. We can thus assume that the object is stationary. The gradient of the graph is thus 0 which means that the speed is also  0 m/s from 0 s to t s. The speed-time graph would thus be as shown in fig 4 below.

Fig 3

 

Fig 4

3. Fig 5 below is a distance-time graph that shows an objects moving and the distance travelled is increasing.

The gradient of the graph is the speed. From the graph the gradient is a constant value but is not zero. Hence the speed-time graph is as shown in fig 6.

Fig 5

Fig 6

4. In fig 7 below we have a distance-time graph that shows an object that is moving as a result the distance travelled is increasing.

As we know the gradient of the graph is the speed and since the gradient is increasing then is it means that the speed is increasing. Hence the speed-time graph of the object is as shown in fig 8.

Fig  7

Fig 8

Distance-time graph

The ability to draw graphs and to obtain information from them is one of the most important skills that a physicist needs to develop. You are also able to extract information more easily using graphs.

Example 1

A boy starts to walk at t = 0 s and walks from point A to point B a distance of 100 m for 10 s . He then stops and walk back towards his starting point in a time of 10 s.  The motion is as shown in fig 1.

 

Fig 1

If you use a table to present this information then it would be as follows:

time /s	Distance travelled /m
0	0
10	100
20	200

Table 1

After 10 s the boy has walked a distance of 100 m. And after 20 s the boy has walked a distance of 200 m( 100 m from A to B and another 100 m from point B to A).

Hence we can plot this on a graph as shown in fig 2 below.

Fig 2

As you can see from this graph the different coordinates will the give you the distance that the boy has walked after a particular time.

Now that you can plot the motion of an object on a graph. Let us see what you can do with a distance-time graph.

You would remember that the speed of an object is the rate of change of distance with time and that it can be calculated using the following equation

speed = distance travelled / time taken

With the distance-time graph the speed of an object at a particular time is the gradient of the line at that particular point.

Now what is the speed of he object at 5 s?

You will have to determine the gradient of the line at 5 s.

Hence the two coordinates that can be used are

(0,0)  and (10,100)

The gradient is thus

gradient = (y1 –y2)/(x1-x2)

                   = (0-100)/(0-10)

                   = –100/-10

                   10

Hence since the gradient a distance-time graph is the speed

speed at 5 s = 10 m/s