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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.

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usmanbiu

Answer:

9.9 s

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) = [tex]\frac{fk}{m}[/tex]
  • acceleration is also the first differential of velocity with respect to time,

       a =[tex]\frac{dv}{dt}[/tex]

  • therefore acceleration (a) = [tex]\frac{fk}{m}[/tex] = [tex]\frac{dv}{dt}[/tex]
  • recall that fk = 70v

        (a) = [tex]\frac{dv}{dt}[/tex] = [tex]\frac{fk}{m}[/tex] = [tex]\frac{70v}{m}[/tex]

        (a) = [tex]\frac{dv}{dt}[/tex] = [tex]\frac{70v}{m}[/tex]

  • integrating both side of the equatin we have

        [tex]\int\limits^v_v₀ {\frac{v}{v₀} } \, = \int\limits^t_0 {\frac{70}{m} } \, t[/tex]

        [tex]ln(\frac{v}{v₀}) = (\frac{70}{m}) t[/tex]

        t =  [tex]\frac{m}{70} x ln(\frac{v}{v₀})[/tex]          

  • the time required for the boat to slow down = t =  [tex]\frac{1000}{70} x ln(\frac{45}{90})[/tex] = - 9.9 s = 9.9 s  

       

       

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