Third equation of motion

We have seen in two earlier post the  first equation of motion and the second equation of motion. If you have not read these two post I would suggest you do so before continuing.

The third equation is derived from the first and the second equations. The mathematics is straight forward as I will show below.

To derive the third law we are going to use the first equation of motion as shown below

v = u +at                             Equation 1

and this second equation that we used when we derived the second equation. If you have forget check the post on the second equation of motion.

s = ¹/₂( v + u)*t                      Equation 2

Now if we change the first equation of motion we will have

                           Equation 3

We can then substitute equation 3 above in equation 2 to obtain

Multiplying on both side by 2a

   Simplifying further

Removing brackets

Simplifying further

So this is the derivation for the third equation of motion.

Example A car travelling at a speed of 10 m/s brakes with a deceleration of 2 m/s² to a stop. Calculate the distance travelled by the car during the deceleration.

To answer this question we first have to identify the different variables.

distance travelled s = ???

initial speed u = 10 m/s

final speed v = o m/s

acceleration = - 2 m/s²

Note that the acceleration is negative since deceleration involves a decrease in speed hence the number is positive.

We now need to use the third equation of motion as shown below.

Substituting the different values in the equation

2*-2*s = o² - 10²

-4s = –100

s =25 m

Hence the distance travelled by the object during the deceleration is 25 m

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.

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