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