2

enter image description here

Since this question covers more aspects of differentiation and mathematical manipulation than kinematics, I am posting it here.

My attempt : On differentiating position(x) once, we get velocity(v) and on differentiating position twice, we get acceleration(A).

Therefore, $xv=at+b$ and $A x^1/3= (vx) (x-vx) $ are the two equations I got. But, I can't seem to express acceleration completely in terms of position, as given in the options below.

Are the options wrong or am I missing something?

Arishta
  • 948
  • I have $v^2+Ax=a$ so that $A={a-v^2 \over x}$ does it agree with your expression? – N74 May 27 '17 at 10:01

1 Answers1

1

Assuming $a,b,c$ constant: $$x=\sqrt{at^2+2bt+c}$$ $$v={at+b\over \sqrt{at^2+2bt+c}}$$ $$A={a\sqrt{at^2+2bt+c} -{(at+b)^2\over \sqrt{at^2+2bt+c}} \over at^2+2bt+c}$$ $$A={a(at^2+2bt+c)-(at+b) ^2\over(\sqrt{at^2+2bt+c} )^3 }$$ $$A={ac-b ^2\over(\sqrt{at^2+2bt+c} )^3 }= {ac-b ^2\over x^3 }$$

N74
  • 2,479