If x is the displacement,v is the velocity,an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.Now,differentiate with respect to $x$:$\frac{d^2t}{dx^2} = \frac{d}{dx}

Vedclass · Accepted Answer

$f = -v^3 \frac{d^2t}{dx^2}$

Vedclass · Answer

(B) We know that velocity $v = \frac{dx}{dt}$,so $\frac{dt}{dx} = \frac{1}{v}$.Now,differentiate with respect to $x$:$\frac{d^2t}{dx^2} = \frac{d}{dx} \left( \frac{1}{v} \right) = -\frac{1}{v^2} \frac{dv}{dx}$.Using the chain rule,$\frac{dv}{dx} = \frac{dv}{dt} \cdot \frac{dt}{dx} = f \cdot \frac{1}{v} = \frac{f}{v}$.Substituting this back into the expression:$\frac{d^2t}{dx^2} = -\frac{1}{v^2} \left( \frac{f}{v} \right) = -\frac{f}{v^3}$.Rearranging the terms,we get $f = -v^3 \frac{d^2t}{dx^2}$.

If $x$ is the displacement,$v$ is the velocity,and $f$ is the acceleration of a particle at time $t$,which of the following is true?

Similar Questions

If $y=a \sin ^3 t$ and $x=a \cos ^3 t$,then $\frac{d y}{d x}$ at $t=\frac{3 \pi}{4}$ is

The equation of the normal to the curve $x=\theta+\sin \theta, y=1+\cos \theta$ at $\theta=\frac{\pi}{2}$ is

The derivative of $\sec ^{-1}\left(\frac{1}{2 x^2-1}\right)$ with respect to $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ equals

If $x = at^2$ and $y = 2at$,then $\frac{dy}{dx} = $ . . . . . . .

If $x=t^2+t+1$ and $y=\sin \left(\frac{t \pi}{2}\right)+\cos \left(\frac{t \pi}{2}\right)$,then find the value of $\frac{dy}{dx}$ at $t=1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module