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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

Book Free Demo
Position-time relation:
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.
 
Fig86.svg
Velocity time graph
 
Distance (\(s\)) travelled by the object,

s=areaOABC(atrapezium)=areaoftherectangleOADC+areaofthetriangleABD=OA×OC+12 (AD×BD)
 
Substituting \(OA\) \(=\) \(u\), \(OC\) \(=\) \(AD\) \(=\) \(t\) and \(BD\) \(=\) \(at\), we get

s=u×t+12t×ats=ut+12at2
Position-velocity relation:
The area enclosed within the trapezium \(OABC\) gives the distance (\(s\)) travelled by the object in time (\(t\)), moving under uniform acceleration (\(a\)).
 
s=areaofthetrapeziumOABCs=OA+BC×OC2
 
Substituting \(OA\) \(=\) \(u\), \(BC\) \(=\) \(v\) and \(OC\) \(=\) \(t\), we get
\(\)
s=(u+v)t2(a)

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

t=vua(b)

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

s=(v+u)(vu)2a
 
(or)
 
2as=v2u2v2=u2+2as