## virtus 3 years ago if v =+/-(x-1) however when x=0m, v= 1m/s >0 why do we take v = -(x-1)

1. ash2326

@virtus do you have a graph along with this?

2. virtus

nope i do not have a graph sorry

3. ash2326

@virtus Is this the full question?

4. virtus

a particle moves along a straight line s that its acceleration is given by a =x-1 where x is its displacement from the origin. Initially, the particles ias at the origin and has velocity of 1m/s

5. virtus

find x as a function of t and describe the motion of the particles

6. ash2326

OK, so we have $a=x-1$ a=acceleration we are given that at t=0, velocity=1m/s , x=0 Do you agree with this?

7. sauravshakya

@ash2326 I dont think x is always 0 when t=0

8. ash2326

Here it's given that "Initially, the particles ias at the origin and has velocity of 1m/s"

9. sauravshakya

oh ya....

10. ash2326

@virtus ??

11. virtus

ok....

12. ash2326

We know that acceleration a $a=\frac{d^2x}{dt^2}$ so we get $\frac{d^2x}{dt^2}=x-1$

13. ash2326

Do you understand this?

14. virtus

yeah

15. ash2326

It's better we use differential equations to solve this Do you know differential equations?

16. ash2326

@virtus ??

17. virtus

no

18. ash2326

we have $a=x-1$ we know $a=\frac {dv}{dt}$ so $\frac{dv}{dt}=x-1$ where v = velocity But here the function is in terms of x, so we'll have to change variables we know that $$v=\frac {dx}{dt}$$ $\frac{dv}{dt}=x-1$ multiplying and dividing left side by dx we get $\frac{dv\times dx}{dt\times dx}=x-1$ or $\frac{dv}{dx}\times\frac {dx}{dt}=x-1$ we know that $$v=\frac {dx}{dt}$$ so we get $\frac{dv}{dx}\times v=x-1$ do you get these steps @virtus ??

19. virtus

thank you i understand

20. ash2326

it's not over yet, still we have some work. Are you here?