Tricky mechanics question

Show that if the acceleration of the particle at a distance $x$ metres from the origin is given by $\displaystyle \frac{1}{100}(x+100) ms^{-1}$, then the particle can start with a velocity of $1 ms^{-1}$ at $O$ and can have a velocity of $3 ms^{-1}$ at a distance of $20m$ from $O$. Show however that it would take 11s to cover this distance.

I have no clue how to go about this!

Write $a=\frac{dv}{dt}=v\frac{dv}{dx}$ and integrate twice.

$\displaystyle a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$

Is this how you get the required form of $a$?

$\displaystyle a = \frac{dv}{dt} = \frac{dx}{dt} \frac{dv}{dx} = v \frac{dv}{dx}$

Is this how you get the required form of $a$?

They've told you what a is!

This form of the chain rule enables you to set up a DE which you can solve for v (and x if necessary) and plug in any boundary conditions.

Was the question actually written the way you typed it - it seems to be phrased very strangely i.e. "can start" and "can have"?

Show that if the acceleration of the particle at a distance $x$ metres from the origin is given by $\displaystyle \frac{1}{100}(x+100) ms^{-1}$, then the particle can start with a velocity of $1 ms^{-1}$ at $O$ and can have a velocity of $3 ms^{-1}$ at a distance of $20m$ from $O$. Show however that it would take 11s to cover this distance.

I have no clue how to go about this!

Just pointing out that you have a unit error in the first line. Acceleration should have $ms^{-2}$.

They've told you what a is!

This form of the chain rule enables you to set up a DE which you can solve for v (and x if necessary) and plug in any boundary conditions.

Was the question actually written the way you typed it - it seems to be phrased very strangely i.e. "can start" and "can have"?

$\displaystyle v \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{1}{100}(x+100)$


$\displaystyle \Rightarrow \frac{v^2}{2} = \frac{1}{100}\left(\frac{x^2}{2} + 100x\right) +c$

Using (0,1) yields c = 1.

$v^2 = \frac{x^2}{100} + 2x + 1$, but this doesn't work out. What am I doing wrong?

$\displaystyle v \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{1}{100}(x+100)$


$\displaystyle \Rightarrow \frac{v^2}{2} = \frac{1}{100}\left(\frac{x^2}{2} + 100x\right) +c$

Using (0,1) yields c = 1.

$v^2 = \frac{x^2}{100} + 2x + 1$, but this doesn't work out. What am I doing wrong?

$\displaystyle v \frac{\mathrm{d} v}{\mathrm{d} x} = \frac{1}{100}(x+100)$


$\displaystyle \Rightarrow \frac{v^2}{2} = \frac{1}{100}\left(\frac{x^2}{2} + 100x\right) +c$

Using (0,1) yields c = 1.

$v^2 = \frac{x^2}{100} + 2x + 1$, but this doesn't work out. What am I doing wrong?

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The question gives the desired answers if the original formula for the acceleration was:

$\displaystyle \frac{1}{100}(x+10)\; ms^{-2}$

Have you given us the complete question?

Can you scan or upload an image of the original? The wording is so strange I'm just suspicious we're missing something!

Is this from an A level paper or a textbook?