14. Second and third equation of motion

Assume an object covering a distance (\(s\)) in time (\(t\)) in an uniform acceleration (\(a\)). The area enclosed within \(OABC\) gives the distance travelled by the object.

 

 

Distance (\(s\)) travelled by the object,

 

Substituting \(OA\) \(=\) \(u\), \(OC\) \(=\) \(AD\) \(=\) \(t\) and \(BD\) \(=\) \(at\), we get

The area enclosed within the trapezium \(OABC\) gives the distance (\(s\)) travelled by the object in time (\(t\)), moving under uniform acceleration (\(a\)).

 

 

Substituting \(OA\) \(=\) \(u\), \(BC\) \(=\) \(v\) and \(OC\) \(=\) \(t\), we get
\(\)

From the velocity-time relation (from the second equation of motion),

$t=\frac{v-u}{a}\to (b)$

Substituting equation (b) on (a) we get,

$s=\frac{(v+u)(v-u)}{2a}$

 

(or)