- #1

Bill_Nye_Fan

- 31

- 2

## Homework Statement

*A body of mass 1 kg moves so that at time ##t## seconds, its displacement is ##x## meters from a fixed origin. The body is acted upon by a force of ##\frac{\sqrt{x-1}}{v^2}## Newtons, where ##v## is the velocity in m/s. The body is initially at rest 1 meter North of the origin. Express ##x## in terms of ##t## and hence find the point in time when the displacement and acceleration have the same magnitude. Express all answers correct to 3 significant figures.*

## Homework Equations

Using calculus to convert between displacement, velocity and acceleration

##a=\frac{dv}{dt}=\frac{d^2x}{dt^2}=v\cdot \frac{dv}{dx}##

##v=\frac{dx}{dt}##

##a=f\left(x\right),\ \frac{d^2x}{dt^2}=f\left(x\right),\ \left(\frac{dx}{dt}\right)^2=2\int _{ }^{ }f\left(x\right)dx,\ v^2=2\int _{ }^{ }f\left(x\right)dx##

## The Attempt at a Solution

##a=\frac{\sqrt{x-1}}{v^2}##, ##v^2=\frac{\sqrt{x-1}}{a}## (equation 1)

Using the last rule in the relevant equations:

##v^2=2\int _{ }^{ }\frac{\sqrt{x-1}}{v^2}dx## (equation 2)

Substituting the first equation into the second:

##\frac{\sqrt{x-1}}{a}=2\int _{ }^{ }\frac{\sqrt{x-1}}{\left(\frac{\sqrt{x-1}}{a}\right)}dx=2\int _{ }^{ }adx##

##\int _{ }^{ }adx=\frac{\sqrt{x-1}}{2a}##

##a=\frac{d\left(\frac{\sqrt{x-1}}{2a}\right)}{dx}##

##a=\frac{\left(\frac{a}{\sqrt{x-1}}-2\sqrt{x-1}\cdot \frac{da}{dx}\right)}{4a^2}##

##4a^3=\frac{a}{\sqrt{x-1}}-2\sqrt{x-1}\cdot \frac{da}{dx}##

##-2\sqrt{x-1}\cdot \frac{da}{dx}=4a^3=\frac{a}{\sqrt{x-1}}##

##\frac{da}{dx}=\frac{a}{2\left(x-1\right)}-\frac{2a^3}{\sqrt{x-1}}##

From here I'm not sure how to simplify this further. Any help is greatly appreciated :)