A motor-boat of mass m moves along a lake with velocity `v_(0)`. At the moment t = 0 the engine of the boat is shut down. Assuming the resistance of the particle to-be proportional to the velocity of the boat F = - rv, find:
(a) How long the motor boat moved with the shut down engine,
(b) (b) The velocity of the motor boat as a function of the distance covered with the shutdown engine, as well as total distance covered till the complete stop.
(c ) The mean velocity of the motor boat over the time interval (beginning with the moment t=0), during which its velocity decreases `eta` times.

Explanation

(a) From the problem `vecF=-vec(rv)` so `m(dvecv)/(dt)=-vec(rv)`
Thus `m(dv)/(dt)=-rv` [as `dvecvuarrdarrvecv`]
or, `(dv)/(v)=-r/mdt`
On integrating `1nv=-r/mt+C`
But at `t=0`, `v=v_0`, so, `C=1nv_0`
or, `1n(v)/(v_0)=-r/mt` or, `v=v_0e^(-r/mt)`
Thus for `trarroo`, `v=0`
(b) We have `m(dv)/(dt)=-rv` so `dv=(-r)/(m)ds`
Integrating within the given limits to obtain `v(s)`:
or, `underset(v_0)overset(v)intdv=-r/m underset(0)overset(s)intds` or `v=v_0-(rs)/(m)` (1)
Thus for `v=0`, `s=s_(t otal)=(mv_0)/(r)`
(c) Let we have `(mdv)/(v)=-rv` or `(dv)/(v)=(-r)/(m)dt`
or, `underset(0)overset(v_0//eta)int (dv)/(v)=(-r)/(m)underset(0)overset(t)int dt`, or, `1n(v_0)/(etav_0)=-r/mt`
So `t=(-m1n(1//eta))/(r)=(m1neta)/(r)`
Now, average velocity over this time interval,
`lt v gt =(int vdt)/(int dt)=(underset(0)overset(m/r 1 n eta)intv_0e^(-eta/mdt))/(m/r1n eta)=(v_0(eta-1))/(eta1eta)`