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Variable acceleration problem

knowledge required: a= dv/dt=v dv/ds

Particle moves along straight line with a = 5-2v v=0, t=0
show that

t = \frac {1}{2} ln (\frac{5}{5-2v})s

and hence express v as a function of t
show that the velocity of the particle approaches a max value and find this max value

(edited 11 years ago)

Reply 1

dantheman1261

Could you post your attempts thus far?

Edit: Are you sure the question information is correct? (Think the 's' in the solution shouldn't be there)

(edited 11 years ago)

Reply 2

I am Ace

aah, only now I realise that that s will be for seconds! Let me try again and post my solution

Reply 3

I am Ace

Is there a way of saving the latex tags? Instead of having to type them out or copy and paste them?
Anyway,

\frac{dv}{dt} =5-2v

t+c'=\int\frac{1}{5-2v}dv

t+c'=\frac{-1}{2}\int\frac{1}{u}du

t+c'=\frac{-1}{2}ln(5-2v)

c=\frac{-1}{2}ln(5-2(0))=\frac{-1}{2}ln5

t=\frac{-1}{2}ln(5-2v)+\frac{1}{2}ln5

Reply 4

FireGarden

That's the same as what they want, but obviously not simplified. It's just laws of logs from there though.

Reply 5

I am Ace

I don't see it,
I end up with

t=\frac{1}{2}ln\frac{5}{5-2v}-\frac{1}{2}

Reply 6

FireGarden

Unparseable latex formula:

[br][br]t = -\dfrac{1}{2} ln(5-2v) + \dfrac{1}{2}ln5 \\[br][br][br]= \dfrac{1}{2}( ln\dfrac{1}{5-2v} + ln5)\\[br][br][br]=\dfrac{1}{2} ln \dfrac{5}{5-2v}[br][br]

(edited 11 years ago)

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Variable acceleration problem

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