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

1. ash2326 Group Title

@virtus do you have a graph along with this?

2. virtus Group Title

nope i do not have a graph sorry

3. ash2326 Group Title

@virtus Is this the full question?

4. virtus Group Title

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 Group Title

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

6. ash2326 Group Title

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 Group Title

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

8. ash2326 Group Title

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

9. sauravshakya Group Title

oh ya....

10. ash2326 Group Title

@virtus ??

11. virtus Group Title

ok....

12. ash2326 Group Title

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

13. ash2326 Group Title

Do you understand this?

14. virtus Group Title

yeah

15. ash2326 Group Title

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

16. ash2326 Group Title

@virtus ??

17. virtus Group Title

no

18. ash2326 Group Title

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 Group Title

thank you i understand

20. ash2326 Group Title

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