With the help of a graph,
derive the relation v^2-u^2=2as
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with with the help of a graph drive the relative V2– U2 =2as
U = velocity at time t1
V= velocity at time t2
a: uniform acceleration of the body along the the straight line
displacement covered during the time interval t2 -–t1 area ab t1 t2
S area of triangle ABA + area of a rectangle AA = t1 t1 area of trapezium.
S =1/2* Sum of parallel side x perpendicular distance .
S = 1/2 (v+u) t ....( Equation)
from the first equation for motion V= u+at ,
t=v–u/a
substituting in equation 1 we get
S=1/2 ( v+u) ( v– u)/a
or , V2 – U2 –2as
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