A boat moves through the water with two forces acting on it. One is a 2.06 103 N forward push by a motor, and the other is a 1.75 103 N resistive force due to the water.
a)What is the acceleration of the 1300 kg boat?
b)If it starts from rest, how far will it move in 11 s?
c)hat will its speed be at the end of its time interval?
F(net) = F-F(res) =(2.06-1.75)•10³=0.31•10³=310 N
a=F(net)/m= 310/1300=0.24 m/s²
s=at²/2=…
v=at = …
To answer the given questions, we need to make use of Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
a) To find the acceleration of the boat, we need to calculate the net force acting on it. The net force can be obtained by subtracting the resistive force due to water from the forward push by the motor:
Net force = Forward push - Resistive force
Net force = 2.06 * 10^3 N - 1.75 * 10^3 N
Net force = 310 N
Next, using Newton's second law, we can find the acceleration:
Acceleration = Net force / mass
Acceleration = 310 N / 1300 kg
Acceleration ≈ 0.238 m/s^2
Therefore, the acceleration of the boat is approximately 0.238 m/s^2.
b) To calculate how far the boat will move in 11 seconds, we can use the equation of motion:
Distance = Initial velocity * time + (1/2) * acceleration * time^2
Since the boat starts from rest, the initial velocity is 0:
Distance = (1/2) * acceleration * time^2
Distance = (1/2) * 0.238 m/s^2 * (11 s)^2
Distance ≈ 14.885 m
Therefore, the boat will move approximately 14.885 meters in 11 seconds.
c) To find the speed of the boat at the end of the time interval, we can use the equation of motion:
Final velocity = Initial velocity + acceleration * time
Since the boat starts from rest, the initial velocity is 0:
Final velocity = 0 + 0.238 m/s^2 * 11 s
Final velocity ≈ 2.618 m/s
Therefore, the speed of the boat at the end of the time interval is approximately 2.618 m/s.