Question

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

Answer 1

@virtus do you have a graph along with this?

Answer 2

nope i do not have a graph sorry

Answer 3

@virtus Is this the full question?

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

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

Answer 8

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

Answer 9

oh ya....

Answer 10

@virtus ??

Answer 11

ok....

Answer 12

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

Answer 13

Do you understand this?

Answer 14

yeah

Answer 15

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

Answer 16

@virtus ??

Answer 17

no

Answer 18

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 ??

Answer 19

thank you i understand

Answer 20

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