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
my attempt:
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 this is as far as i can , i just don't get how the answer become xt = √(vv0)t so could someone help me ?
ps: i'm sorry if this quite messy since i didnt use latex , i confuse how to use it.
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
my attempt:
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 this is as far as i can , i just don't get how the answer become xt = √(vv0)t so could someone help me ?
ps: i'm sorry if this quite messy since i didnt use latex , i confuse how to use it.