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Is there something special when an object moves according to the law a=k*sqrt(v) ? first have a look at the question , http://in.answers.yahoo.com/question/ind… i now understand the intuitive reason , 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 explain why the derivative comes out to be constant ? if a body travels according to the law a=k*sqrt(v) then also da/dt comes out to be constant. i was discussing this with my sister, and she said this has something to do with the fact that derivative of parabolic function,y=x^2 is proportional to the x- coordinate of the point, and because y=sqrt*x is its inverse function hence something happens ( i don't know what to write now) plz answer if you understood what i'm trying to say...


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


first have a look at the question ,

http://in.answers.yahoo.com/question/ind…
i now understand the intuitive reason ,
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 explain why the derivative comes out to be constant ? if a body travels according to the law a=k*sqrt(v) then also da/dt comes out to be constant.
i was discussing this with my sister, and she said this has something to do with the fact that derivative of parabolic function,y=x^2 is proportional to the x- coordinate of the point, and because y=sqrt*x is its inverse function hence something happens ( i don't know what to write now)
plz answer if you understood what i'm trying to say...

Grade:12

1 Answers

shubham tomar
31 Points
10 years ago

the http://............ is a clickable link.

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