Mechanics limits query. (1 Viewer)

[joshlols]

From Cambridge 4unit Diagnostic Test 7, Question 4.
A particle of mass m moves in a horizontal straight line away from a fixed point O in a line. The particle is resisted by a force mkv^(3/2), where k is a positive constant and v is the speed. When t = 0, v = u > 0. Show that the particle is never brought to rest and that its distance from O is at most [2sqrt(u)]/k.

The second part looks like this:

x = 2/k[sqrt(u) - sqrt(v)]

Which makes sense to me that x -> [2sqrt(u)]/k IF v -> 0 but am wondering how or why that would happen.

Edit: Nevermind on the first part of the question.

 

[vds700]

From Cambridge 4unit Diagnostic Test 7, Question 4.
A particle of mass m moves in a horizontal straight line away from a fixed point O in a line. The particle is resisted by a force mkv^(3/2), where k is a positive constant and v is the speed. When t = 0, v = u > 0. Show that the particle is never brought to rest and that its distance from O is at most [2sqrt(u)]/k.

The second part looks like this:

x = 2/k[sqrt(u) - sqrt(v)]

Which makes sense to me that x -> [2sqrt(u)]/k IF v -> 0 but am wondering how or why that would happen.

Edit: Nevermind on the first part of the question.

R = ma = -mkv^3/2)
a = -kv^(3/2)
dv/dt = -kv^(3/2) if u inverrt this and integrate, u will get an equation for time

t = 2k[(1/sqrtv) - (1/sqrtu)]
if u put v = 0, t will be infinity. So it doesn't come to rest UNTIL t = infinty.

Put v = 0 into your equation for displacement and you'll see that it approaches 2/k[sqrt(u)] as t approaches infinity.

Poor wording of the question IMO