Is there something special when an object moves according to the law v=a*sqrt(s) ? v -velocity a-constant s-displacement actually, the acceleration for this type of motion comes out to be constant. dv/dt=dv/ds*ds/dt = [a/2*sqrt(s)]*a.sqrt(s)=a^2/2 can you intuitively explain why this happens ?

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vikas askiitian expert · Answer

v=ds/dt=Ksqrt(s) (k=a^2/2) ds/sqrt(s)=kdt 2sqrts=kt+c s=(kt+c)^2 /2 it means graph between time and displacement is parabola......

Askiitians_Expert Yagyadutt · Answer

Hiii shubham

v = a root(x)

dx/dt = a.root(x)

dx/a.root(x) = a.dt

Integrate it ...

2root(x) = a .t

or x = a^2.t^2/4

x = kt^2

This is a parabolic path with respect to time..

d^2x/dt^2 = 2k ( constant )

So this is just a projectile motion with object starting from rest...moved with constant acceleration 2k

Regards

Yagya

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Is there something special when an object moves according to the law v=asqrt(s) ? v -velocity a-constant s-displacement actually, the acceleration for this type of motion comes out to be constant. dv/dt=dv/dsds/dt = [a/2sqrt(s)]a.sqrt(s)=a^2/2 can you intuitively explain why this happens ?

Is there something special when an object moves according to the law v=a*sqrt(s) ?

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Is there something special when an object moves according to the law v=a*sqrt(s) ? v -velocity a-constant s-displacement actually, the acceleration for this type of motion comes out to be constant. dv/dt=dv/ds*ds/dt = [a/2*sqrt(s)]*a.sqrt(s)=a^2/2 can you intuitively explain why this happens ?

Is there something special when an object moves according to the law v=a*sqrt(s) ?

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Is there something special when an object moves according to the law v=asqrt(s) ? v -velocity a-constant s-displacement actually, the acceleration for this type of motion comes out to be constant. dv/dt=dv/dsds/dt = [a/2sqrt(s)]a.sqrt(s)=a^2/2 can you intuitively explain why this happens ?