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.

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

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  p_1b is the linear momentum of object 1 before the collision and p_2b

is the linear momentum of  object 2 before the collision then

the sum of linear momentum = p_1b +p_2b

= 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  p_1a is the linear momentum of object 1 after the collision and p_2a

is the linear momentum of  object 2 after the collision then

the sum of linear momentum = p_1a +p_2a

= 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

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

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 = p_ic  +p_ib

where p_ic is the initial momentum of the cannon and p_ib is the initial momentum of the ball.

After firing the cannon ball

sum of linear momentum after firing= p_fc  +p_fb

where p_fc is the final momentum of the cannon and p_fb 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 = p_ic  +p_ib

= 1000*0 + 15*0

= 0 kg m/s

sum of linear momentum after firing= p_fc  +p_fb

=1000*v_rc +15*150

=1000*v_rc +2250

where is the v_rc 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*v_rc +2250

0 – 2250 = 1000*v_rc

-2250 = 1000*v_rc

v_rc = –2.25 m/s

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

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  = (v_f – v_i)/t

Rearranging the equation will give you

v_f = v_i + 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

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.

What is linear momentum?

Linear momentum is a physical quantity that depends on the mass and the velocity of the object.

Fig 1 below shows an object that has mass m and is moving with a velocity v.

 

The linear momentum which is denoted by the symbol p and is calculated using the equation below

 

linear momentum  = mass of object  x velocity of object

p = mv

Hence the unit for linear momentum is kg m s^-1

The linear momentum is thus a vector quantity since it is the product of a scalar quantity (the mass) and the vector quantity (the velocity).

We can thus say that the linear momentum of a body is defined as the product of the mass of that body and its velocity

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.

Kinetic energy

Kinetic energy is one if the eight forms of energy. It is simply the energy that a body possesses due to its motion.

The kinetic energy of a body is simply calculated using the equation

Kinetic energy Ek = ½ mv²

Where m is the mass of the mass of the object in kg
v is the velocity of the object in ms^-1

As you can see the kinetic energy of the object depends on both the velocity and the mass of the object. So it is possible for an elephant with a large mass to have the same kinetic energy as a small object moving at a high velocity.

Now let us see an example where the kinetic energy of an object is calculated.

Example 1

An elephant of mass _{3.00 x 10}³ kg is moving at a velocity of 3.00 ms^-1. What is the kinetic energy of the elephant?

So let us recall that the kinetic energy of the elephant is calculated using the equation

Ek = ½ mv²

Hence the kinetic energy Ek = ½ mv²
= ½(_{3.00* 10}³)*(3.00)²
= 13500 J
= _{1.35 *10}⁴ J

Now you can be given the kinetic energy and be asked to calculate the mass or the velocity. Let us look at a second example.

Example 2

An object has a mass of 4.0 kg and a kinetic energy of 16 J. Determine the velocity of the object.

Ek = ½ mv²
16 = ½ (4.0)*v²

Making v the subject of formula

v² = (16 *2)/4.0
v = (16*2/4.0)^½
v = 2.8 ms^-1

Now to test your newly acquired you can do the following questions

1. Calculate the kinetic energy possessed by an aeroplane of mass _{3.4 * 10}⁵ kg flying at 110 ms.
2. If a body has _{3.4 x 10}⁵ J of kinetic energy and mass _{3.2* 10}² kg, what is its velocity?