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## Homework Statement

acceleration of moving particle is described by

a=-kv^1,5 where k is a constant. if the condition when t=0 is v=v0 and x=0 prove that xt = √(vv0).t

## Homework Equations

dv/dt=a, dx/dt=v

## The Attempt at a Solution

dv/dt=a

dv/v^1,5=-k dt

v^-1,5 dv = -k dt ← integrating both sides

-2(v^-½)= -kt + C or

(v^-½) = ½kt + C then plug the initial condition : since when t = 0 v = v0 this turned into

(v0^-½) = ½k(0) + C

so the value of C is equal to v0^-½ continuing by insert the value of C into general equation:

v^-½= ½kt + v0^-½ and for value of v :

v = (½kt + v0^-½)^-²

then move into x

dx/dt = v

dx = v dt

dx = (½kt + v0^-½)^-² dt ← integrating both sides

x = -2/k (½kt + v0^-½)^-¹ + C

the given condition is when t is equal to zero x is also zero thus,

0 = -2/k ( 0 + v0^-½)^-1 + C then re-prashe

(v0^-½)^-1 into v0^½

it would became

0 = -2/k v0^½ + C and the value of C is: 2/k v0½

and i go back to general equation

x = -2/k (½kt + v0^-½)^-¹ + 2/k v0½

the question demand for xt = √(vv0)t and i somewhat know that (½kt + v0^-½)^-¹ is equal to √v

back at

v^-½= ½kt + v0^-½ so this implies v^½ or √v is equal to (½kt + v0^-½)^-¹ so i rewrite the equation into

x = -2/k v½ + 2/k v0½

but, i cant go any further than this i just cannot understand how xt=√(vv0)t obtained, and i did not thoughts that i made mistakes in my integrating process, so can anyone guide me to past this ? and i'm sorry i don't know the suitable title for this, since i not familiar with physics terms i english, so please excuse me.