The force of resistance encountered by water on a motor boat of mass `m` going in still water with velocity `v` is proportional to the velocity `vdot` At `t=0` when its velocity is `v_0,` then engine shuts off. Find an expression for the position of motor boat at time `t` and also the distance travelled by the boat before it comes to rest. Take the proportionality constant as `k > 0.`

Explanation

The resistance force opposing the
motion = m `xx` acceleration `=(dv)/(dt)`
Hence, differential equation is `m(dv)/(dt) = -kv`
or `(dv)/(v) = -k/m dt`
Integrating, we get ln v`=-k/mt+c`
At t=0, `v=v_(0)`, Hence, `c="ln"v_(0)`
`therefore "ln"v/v_(0) = -k/m.t` or `v=v_(0)e^(-k/mt)`.............(1)
Where v is the velocity at time t.
Now, `(ds)/(dt) = v_(0)e^(-k/mt)`
or `ds= v_(0)e^(-k/mt) +C`
Boat's position at time t is
`s(t) = (-v_(0)m)/(k) e^((-kt)/(m))+c`
If `t=0, s=0` or `c=(v_(0)m)/(k)`
`therefore s(t) = (v_(0)m)/(k) [1-e^((-kt)/(m))]`..............(2)
To find how far the boat goes, we have to find `lim_(t to infty)s(t)` which is `(mv_(0))/(k)`.