Question

A boat moves through the water with two forces acting on it. One is a 1,950-N forward push by the water on the propeller, and the other is a 1,400-N resistive force due to the water around the bow. (a) What is the acceleration of the 1,200-kg boat? m/s 2 (b) If it starts from rest, how far will the boat move in 20.0 s? m (c) What will its velocity be at the end of that time? m/s

Answer 1

Mass of the boat = m = 1200 kg

Forward force on the boat = F₁ = 1950 N

Resistive force on the boat = F₂ = -1400 N (Negative as it is resisting the motion of the boat)

Net force on the boat = F_net

F_net = F₁ + F₂

F_net = 1950 + (-1400)

F_net = 550 N

Acceleration of the boat = a

F_net = ma

550 = (1200)a

a = 0.4583 m/s²

Initial velocity of the boat = V₁ = 0 m/s (Starts from rest)

Velocity of the boat after 20 sec = V₂

Time period = T = 20 sec

V₂ = V₁ + aT

V₂ = 0 + (0.4583)(20)

V₂ = 9.166 m/s

Distance the boat travels in 20 sec = D

D = V₁T + aT²/2

D = (0)(20) + (0.4583)(20)²/2

D = 91.66 m

a) Acceleration of the 1200 kg boat = 0.4583 m/s²

b) Distance the boat will move in 20 sec = 91.66 m

c) Velocity of the boat at the end of 20 sec = 9.166 m/s