Question

A 1000 kg boat is traveling at 90 km/h when its engine is shut off. The magnitude of the frictional force f → k between boat and water is proportional to the speed v of the boat: fk = 70v, where v is in meters per second and fk is in newtons. Find the time required for the boat to slow to 45 km/h.

Answer 1

9.9 s

Step-by-step explanation:

mass (m) = 1000 kg

initial speed (u) = 90 km/h

final speed (v) = 45 km/h

relationship between the speed (v) of the boat and the frictional force (fk) ⇒ fk = 70v

• the acceleration of the system will be given by (a) =
  `(fk)/(m)`

• acceleration is also the first differential of velocity with respect to time,

a =
`(dv)/(dt)`

• therefore acceleration (a) =
  `(fk)/(m)` =
  `(dv)/(dt)`
• recall that fk = 70v

(a) =
`(dv)/(dt)` =
`(fk)/(m)` =
`(70v)/(m)`

(a) =
`(dv)/(dt)` =
`(70v)/(m)`

• integrating both side of the equatin we have

`\int\limits^v_v₀ {(v)/(v₀) } \, = \int\limits^t_0 {(70)/(m) } \, t`

`ln((v)/(v₀)) = ((70)/(m)) t`

t =
`(m)/(70) x ln((v)/(v₀))`

• the time required for the boat to slow down = t =
  `(1000)/(70) x ln((45)/(90))` = - 9.9 s = 9.9 s