suppose-the-acceleration-of-a-particle-moving-along-a-straight-path-is-given-by-

Question

Suppose the acceleration of a particle moving along a straight path is given by a function of position a(x)=−k(1+xL)2a\left(x\right)=-\frac{k}{\left(1+\frac{x}{L}\$$right)^2$$}, where kk and LL are positive constants. Using the Chain Rule, relate dvdt\frac{dv}{dt} and ddx(v2)\frac{d}{dx}\$$left(v^2$$\right), where v=dxdtv=\frac{dx}{dt}.

Accepted Answer

dvdt=12ddx(v2)\frac{dv}{dt}=\frac12\frac{d}{dx}\$$left(v^2$$\right)

Answer

dvdt=2ddx(v2)\frac{dv}{dt}=2\frac{d}{dx}\$$left(v^2$$\right)

Answer

dvdt=ddx(v2)\frac{dv}{dt}=\frac{d}{dx}\$$left(v^2$$\right)

Answer

dvdt=14ddx(v2)\frac{dv}{dt}=\frac14\frac{d}{dx}\$$left(v^2$$\right)