deduce ut+1/2at^2 by graphical method
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the image is attached with the answer if you like it mark it the brainliest
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kanna20:
bro its not kinematics
what
that chapter is also called kinematics(motion in a stright line)
bro
macha anyways thanks
can you please mark it brainliest
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Distance travelled(s)=Area of trapezium OABC
s=(Area of ∆ABD + Area of rectangle OADC)
s=(1/2×AD×BD)+OC×OA ....(1)
From the figure
AD=OC=t
BD=(BC-DC)=(v-u)
=(u+at-u)=at
(since v=u+at)
OA=u
Substituting equation 1, we get
s=(1/2×t×at)+(t×u)
s=(1/2at^2)+(ut)
so s=ut+1/2at^2
This is the second equation of motion
s=(Area of ∆ABD + Area of rectangle OADC)
s=(1/2×AD×BD)+OC×OA ....(1)
From the figure
AD=OC=t
BD=(BC-DC)=(v-u)
=(u+at-u)=at
(since v=u+at)
OA=u
Substituting equation 1, we get
s=(1/2×t×at)+(t×u)
s=(1/2at^2)+(ut)
so s=ut+1/2at^2
This is the second equation of motion
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