A boat moves through the water with two forces acting on it. One is a 2100. N forward push by the water on the propeller, and the other is a 1600. N resistive force due to the water around the bow.
(a) What is the acceleration of the 1200. kg boat?
(b) If it starts from rest, how far will the boat move in 40.0 s?
(c) What will its velocity be at the end of that time?
the net force acting on the boat is 2100 N - 1600 N = 500 N
a) Force = mass * acceleration
500 N = 1200 * a
Solve for a, the acceleration in m/s^2
b) distance = 1/2 * a * t^2
where t is time
distance = 1/2 *a*40^2
Plug in the acceleration from (a) and solve for the distance in m
c) velocity = a * t
v = a * 40.0
Plug in a from part (a), and solve for the velocity in m / s
(a) Well, to find the acceleration of the boat, we need to first find the net force acting on it. The net force is given by the difference between the forward push and the resistive force: 2100 N - 1600 N = 500 N.
Next, we can use Newton's second law, which states that force equals mass times acceleration (F = m * a). Plugging in the values, we have 500 N = 1200 kg * a.
Solving for acceleration (a), we get a = 500 N / 1200 kg ≈ 0.417 m/s².
(b) If the boat starts from rest, we can use the equation of motion d = (1/2) * a * t² to find the distance it will cover in 40.0 seconds. Plugging in the values, we have d = (1/2) * 0.417 m/s² * (40.0 s)².
Simplifying this equation, we get d = 0.5 * 0.417 m/s² * 1600 s² = 333.6 meters.
So, the boat will move approximately 333.6 meters in 40.0 seconds.
(c) To find the velocity of the boat at the end of the given time, we can use the equation v = a * t. Plugging in the values, we have v = 0.417 m/s² * 40.0 s.
Simplifying this equation, we get v = 16.68 m/s.
Therefore, the boat will have a velocity of approximately 16.68 m/s at the end of that time.
(a) To find the acceleration of the boat, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
The net force is the difference between the forward push force and the resistive force:
Net force = Forward push force - Resistive force
= 2100 N - 1600 N
= 500 N
Using the formula F = ma, we can rearrange it to solve for acceleration:
a = F/m
= 500 N / 1200 kg
≈ 0.42 m/s^2
Therefore, the acceleration of the boat is approximately 0.42 m/s^2.
(b) To find the distance the boat will move in 40.0 s, we can use the kinematic equation:
d = v0t + (1/2)at^2
Since the boat starts from rest, the initial velocity (v0) is 0 m/s. Plug in the known values:
d = 0(40.0) + (1/2)(0.42)(40.0)^2
≈ (1/2)(0.42)(1600)
≈ 336 meters
Therefore, the boat will move approximately 336 meters in 40.0 seconds.
(c) To find the velocity of the boat at the end of 40.0 seconds, we can use another kinematic equation:
v = v0 + at
Since the boat starts from rest, the initial velocity (v0) is 0 m/s. Plug in the known values:
v = 0 + (0.42)(40.0)
= 16.8 m/s
Therefore, the velocity of the boat at the end of 40.0 seconds is 16.8 m/s.
To determine the acceleration of the boat, we need to calculate the net force acting on it and then use Newton's second law of motion, F = ma, where F is the net force, m is the mass of the object (boat), and a is the acceleration.
(a) Calculating the net force:
The net force is obtained by subtracting the resistive force (1600 N) from the forward push force (2100 N).
Net force = Forward push force - Resistive force
Net force = 2100 N - 1600 N
Net force = 500 N
Next, we can use Newton's second law to calculate the acceleration:
F = ma
500 N = 1200 kg * a
Rearranging the equation, we find:
a = 500 N / 1200 kg
a ≈ 0.417 m/s²
Therefore, the acceleration of the boat is approximately 0.417 m/s².
(b) To determine the distance the boat will move in 40.0 s, we can use the kinematic equation:
d = v₀t + 0.5at²
Since the boat starts from rest (v₀ = 0 m/s), the equation simplifies to:
d = 0.5at²
Plugging in the values, we have:
d = 0.5 * 0.417 m/s² * (40.0 s)²
Simplifying the equation gives us:
d ≈ 333.6 m
Therefore, the boat will move approximately 333.6 meters in 40.0 seconds.
(c) Finally, to determine the velocity of the boat at the end of 40.0 seconds, we can use another kinematic equation:
v = v₀ + at
Since the boat starts from rest (v₀ = 0 m/s), the equation simplifies to:
v = at
Plugging in the values, we get:
v = 0.417 m/s² * 40.0 s
Simplifying the equation gives us:
v ≈ 16.7 m/s
Therefore, the velocity of the boat at the end of 40.0 seconds will be approximately 16.7 m/s.