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

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

Motion and equations of motion

Distance and displacement

Speed and velocity

Acceleration

Gradient and how to determine it?

Types of distance-time graph

First equations of motion

Second equation of motion

Third equation of motion