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Grade 11General Physics

The acceleration of a particle moving in a straight line varies with displacement as a=2s ,velocity of particle is zero at zero displacement. Find corresponding velocity-displacement relationship

Profile image of Shashank Raut
10 Years agoGrade 11
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1 Answer

Profile image of Riddhish Bhalodia
10 Years ago
a = \frac{dv}{dt} = \frac{dvds}{dtds} = v\frac{dv}{ds} = 2s

\int vdv = 2\int sds
v^2 = 2s^2 + c
and as v=0 at s=0 we have c=0
so
v^2 = 2s^2
is the relationship