Irodov 1.23 confusion: Calculate distance traveled and time elapsed for this decelerating particle

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Homework Statement
A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as shown in book, where a is a positive constant. At the initial moment the velocity of the point is equal to v0. What distance will it traverse before it stops? What time will it take to cover that distance?
Relevant Equations
acceleration=dv/dt=vdv/ds
v=ds/dt
I solved the for time but for distance am quite confused on why the approach im trying doesn't work.

First, let acceleration=w
w=-a√v
w=dv/dt=-a√v
rearranging, √v dv=-a dt

integrating both sides,
v0V∫√v dv=0T∫-a dt
where V is some arbitrary velocity at time T.
we get, (2√V)-(2√v0)=-aT------eqn 1

now we want to find time when V=0
putting V=0,
aT=2√v0
T=(2√v0)/a

now, to find distance, i took equation 1 and rearranged.
2(ds/dt)1/2=-aT+2√v0
squaring both sides,
ds/dt=(2√v0-aT)2/4

integrating,
s=(1/4)∫(2√v0-aT)2 dt where s is distance travelled
let (2√v0-aT)2=u
du/dt=-a
dt=du/-a
putting, (1/4)∫(2√v0-aT)2dt=-(1/4a)∫u2 du
this gives, s=-1/12((2√v0-aT)2)3

putting t=(2√v0)/a, we get s=0 which simply isnt possible and is the wrong answer. Can anyone tell me the mistake in my approach?

p.s. i know how to solve it using w=vdv/dt so please don't give me that solution.
Thanks
 
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anthraxiom said:
Homework Statement: A point moves rectilinearly with deceleration whose modulus depends on the velocity v of the particle as shown in book, where a is a positive constant. At the initial moment the velocity of the point is equal to v0. What distance will it traverse before it stops? What time will it take to cover that distance?
Relevant Equations: acceleration=dv/dt=vdv/ds
v=ds/dt

I solved the for time but for distance am quite confused on why the approach im trying doesn't work.
First, let acceleration=w
w=-a√v
w=dv/dt=-a√v
rearranging, √v dv=-a dt
integrating both sides,
v0V∫√v dv=0T∫-a dt
where V is some arbitrary velocity at time T.
we get, (2√V)-(2√v0)=-aT------eqn 1
now we want to find time when V=0
putting V=0,
aT=2√v0
T=(2√v0)/a
now, to find distance, i took equation 1 and rearranged.
2(ds/dt)1/2=-aT+2√v0
squaring both sides,
ds/dt=(2√v0-aT)2/4
integrating,
s=(1/4)∫(2√v0-aT)2 dt where s is distance travelled
let (2√v0-aT)2=u
du/dt=-a
dt=du/-a
putting, (1/4)∫(2√v0-aT)2dt=-(1/4a)∫u2 du
this gives, s=-1/12((2√v0-aT)2)3
putting t=(2√v0)/a, we get s=0 which simply isnt possible and is the wrong answer. Can anyone tell me the mistake in my approach?
p.s. i know how to solve it using w=vdv/dt so please don't give me that solution.
Thanks
In integrating to get s, you forgot to apply the lower integration limit at t=0.
 
Chestermiller said:
In integrating to get s, you forgot to apply the lower integration limit at t=0.
Oh thank you so much I've been banging my head about this for 3 hours.