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
![clip_image002[4]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhqkfxsHJAyWPTAx6APIKvgApqjxybyiz8aTNq8F6UDyv220AOop3nusPg84mwfhrBYf0YfAQyyKGHcyfxZEn0-2-9JncU36KLwj6i_n9QsBgIwW99GwncW2E1gafFXvzcisd9755b3cUZv/s1600-h/clip_image002%5B4%5D%5B2%5D.gif "clip_image002[4]"){kind=small}
Equation 3We can then substitute equation 3 above in equation 2 to obtain
![clip_image002[10]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgogaMdWswEr1jJv26KfbfFk_ZyXNiikG26rw_gp9pZRwG1idyqN60LbnGijgzjnxacQfHPi5KS8NVuEqQsomKBPBvbxiPTcCELhekrs6y4A6lOb8X3LBFAy6C4EnGSnGrjbUKsBBPZITQM/?imgmax=800 "clip_image002[10]"){kind=small}
Multiplying on both side by 2a
![clip_image002[12]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgeUxYOiIzKjz3J47qx-pnKfBErEmYCxFLqaHQGfF-5Gzg10NH89YJSgjUT8BKOEcfTtsjrXPjqwO00jxAGYZFlfaPQp9zA_x6Cvm14iOyTdfv_2u-yq2GLmx9N3GPLOHwBIMnjPuLrP7t5/s1600-h/clip_image002%5B12%5D%5B2%5D.gif "clip_image002[12]"){kind=small}
Simplifying further
![clip_image002[16]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi39vKAe_1HlI4UCDDkl0b21bJ_xXjiy_qvl6IymzEZmGVL7I0WtlSZ5w8gX50c0Ebuho9mgov-_Q1loBushrSHDxUDXP5_7_Vu5VC1Z8fdz3IVdtB7oxQman3ChJsu1Bk0s8UKXc0LRR3k/s1600-h/clip_image002%5B16%5D%5B2%5D.gif "clip_image002[16]"){kind=small}
Removing brackets
![clip_image002[18]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEiq7YUyD1skFlME7IX4QnJ9z9Sk3mNCJ65IkKDLAY5w3um53vjJKtt1XdLlD0xqCw4mBj_ZPf1ZVuKyp3MEuL95Fe6KTO54DNsxgMMqRNLCT-n8cSwHpuVdX4tNp30LW5EjHBC_9icaHJ_y/s1600-h/clip_image002%5B18%5D%5B2%5D.gif "clip_image002[18]"){kind=small}
Simplifying further
![clip_image002[20]](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEh_Po5sqT2yRoNWAGwdGQJRi-uCWawgV5AMznPHUAjYBqE6iJwWXFRHfw-vbGBYC7MsHGoakaxu56xcvEP2pYEP1Szg5svq0rPFbV5oNzCZhy78qm77E0cISMpzG6Plv57EsOaUBo7Z7Pn-/s1600-h/clip_image002%5B20%5D%5B2%5D.gif "clip_image002[20]"){kind=small}
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 mWe 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 3We 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 mWe 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 3We 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
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