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.

Speed and velocity

After distance and displacement that we have seen earlier, I am now going to talk about speed and velocity. They are quite different and to understand them it is important to understand clearly the difference distance and displacement before going forward.

 

Speed

Speed is a scalar quantity and the unit is the metre per second (m s^-1)

A boy moves from point A and walks a distance of 50 m to the point B in a time of 20 s.

Firstly what is the distance travelled in 1 second?

Yes the boy would walk a distance of 2.5 m in a time of 1 second.

The distance travelled in a time of 1 second is called the speed.

The equation to calculate the speed is shown below.

 

 

Can you determine the speed of the boy?

Hence whenever speed is calculated the distance should be used.

Velocity

Velocity is a vector quantity and the unit is also the metre per second(m s^-1)

Since it is a vector quantity then when it is determined direction is important and hence displacement should be used.

A boy moves from point A to point B performing a displacement of 50 m towards the right as shown in the diagram below.

The following equation is used to determine the velocity.

Hence the velocity of the boy can be calculated as shown below.

Please note the difference between the two. For the speed the distance travelled is used while for the velocity the displacement is used.