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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) ?


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 ?

Grade:12

2 Answers

vikas askiitian expert
509 Points
10 years ago

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
askIITians Faculty 74 Points
10 years ago

 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

askiitians_expert

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