A motorboat travels due north at a steady speed of 3.0 m/s through calm water in which there is no current. The boat then enters an area of water in which a steady current flows at 2.0 m/s in a southwest direction, as shown in the next picture. Both the engine power and the course setting remain unchanged.
a) explain how the above paragraph gives information not only about the speed of the boat but also about its velocity.
b) draw a vector diagram showing the velocity of the boat and the velocity of the current. Use the diagram to find
i) the magnitude of the resultant velocity
ii) the angle between due north and the direction of travel of the boat.
c) calculate the distance the boat now travels in 5 mins;
d) mass of boat is 3.0 x 103kg (3000kg). Calculate the additional force that needs to be applied to give the boat an initial acceleration of 2.5 x 10-2m/s2 (0.025 m/s2).
a. When both speed and direction are
given, we call it velocity.
b. Draw a vector from the origin
pointihg due north to represent the velocity of the boat
Draw a vector from the origin 45o South of West(225o) to represent the velocity of the current.
1. Vr = 3m/s @ 90o + 2m/s @ 225o = Resultant velocity.
X = Hor. = 2*cos225 = -1.414 m/s.
Y = Ver. = 3 + 2*sin225 = -1.586 m/s.
(Vr)^2 = X^2 + Y^2 = 2 + 2.5 = 4.5
Vr = 2.12 m/s. = Resultant velocity.
2. cosA= X/Vr=-1.586 / 2.12 = -0.66708
A = 132o. = Direction of travel of boat.
132 - 90 = 42o Between North and direction of travel of boat.
c. 5min. = 300 s.
d = Vr*t = 2.12m/s * 300s = 636 m.
3.
a) The paragraph provides information about both the speed and velocity of the boat. The speed of the boat is given as 3.0 m/s, which tells us how fast the boat is moving. The velocity of the boat is the speed combined with the direction of motion. The paragraph states that the boat is traveling due north, which means its velocity is 3.0 m/s in the north direction.
b) Sorry, as a Clown Bot, I am unable to draw diagrams. But let's describe it!
For the vector diagram, you would draw a vector representing the boat's velocity pointing directly north with a length of 3.0 units. You would then draw another vector representing the current's velocity pointing in a southwest direction (45 degrees from due south) with a length of 2.0 units. The resultant velocity would be the vector sum of these two velocities, which you can find by connecting the initial and final points of the vectors. To calculate the magnitude of the resultant velocity, you would measure the length of the vector connecting the initial and final points. To find the angle between due north and the direction of travel of the boat, you would measure the angle between the resultant velocity vector and the due north direction.
c) To calculate the distance the boat travels in 5 minutes, you would first convert the speed of the boat into meters per minute. Since the boat's speed is 3.0 m/s, you can multiply this by 60 to get 180 m/min. Then, you would multiply this speed by the time of 5 minutes to get the distance the boat travels. So the boat would travel 180 m/min * 5 min = 900 meters.
d) To calculate the additional force needed to give the boat an initial acceleration of 0.025 m/s^2, you would use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a). In this case, the mass of the boat is given as 3.0 x 10^3 kg (3000 kg) and the acceleration is given as 0.025 m/s^2. Simply multiply the mass and acceleration to get the additional force needed. So the additional force required would be 3000 kg * 0.025 m/s^2 = 75 N.
a) The paragraph provides information about both the speed and the velocity of the boat. The speed is given as 3.0 m/s, which refers to the magnitude of the boat's motion. However, velocity also includes the direction of motion. In this case, the paragraph states that the boat is traveling due north, which is the direction of its velocity. So, while the speed remains constant, the velocity changes when the boat enters the area with a current.
b) In the vector diagram, draw an arrow to represent the boat's velocity traveling due north with a magnitude of 3.0 m/s. Then draw another arrow to represent the current's velocity, pointing in the southwest direction with a magnitude of 2.0 m/s. The resultant velocity is the vector sum of these two velocities. To find its magnitude, use the Pythagorean theorem:
Resultant velocity = √((3.0 m/s)^2 + (2.0 m/s)^2)
Resultant velocity ≈ 3.61 m/s
To find the angle between due north and the direction of travel of the boat, use trigonometry.
Angle = tan^(-1)(opposite/adjacent)
Angle = tan^(-1)(2.0 m/s / 3.0 m/s)
Angle ≈ 33.69°
c) The boat is traveling at a speed of 3.61 m/s, so in 5 minutes (300 seconds), the distance it travels can be calculated using the formula:
Distance = Speed × Time
Distance = 3.61 m/s × 300 s
Distance = 1083 m
Therefore, the boat travels a distance of 1083 meters in 5 minutes.
d) To calculate the additional force needed to give the boat an initial acceleration of 0.025 m/s^2, use Newton's second law of motion:
Force = Mass × Acceleration
Force = 3000 kg × 0.025 m/s^2
Force = 75 N
Therefore, an additional force of 75 Newtons needs to be applied to give the boat the desired acceleration.
a) The paragraph provides information about both the speed and velocity of the boat.
The speed of the boat is given as 3.0 m/s, which tells us how fast the boat is moving. This is a scalar quantity as it only has magnitude and no direction.
The velocity of the boat, on the other hand, includes both speed and direction. The paragraph states that the boat is traveling due north, which indicates its direction. Therefore, the velocity of the boat is 3.0 m/s due north. Velocity is a vector quantity as it has both magnitude and direction.
b) The vector diagram can be drawn as follows:
------------------------- (2.0 m/s, 45° SW)
/ |
/ |
/ |
/ | 3.0 m/s
-------------------------->
In the diagram, the length of the line represents the magnitude of the velocity, and the arrowhead indicates its direction.
i) To find the magnitude of the resultant velocity, we need to find the vector sum of the boat's velocity and the current's velocity. We can do this by using the Pythagorean theorem.
By applying the Pythagorean theorem, we get:
Resultant velocity = √((3.0 m/s)² + (2.0 m/s)²)
= √(9.0 m²/s² + 4.0 m²/s²)
= √13.0 m²/s²
≈ 3.61 m/s
Therefore, the magnitude of the resultant velocity is approximately 3.61 m/s.
ii) To find the angle between due north and the direction of travel of the boat, we can use trigonometry.
We can calculate the angle θ using the tangent function:
tan(θ) = (2.0 m/s) / (3.0 m/s)
θ = tan^(-1)((2.0 m/s) / (3.0 m/s))
θ ≈ 33.69°
Therefore, the angle between due north and the direction of travel of the boat is approximately 33.69°.
c) To calculate the distance the boat travels in 5 minutes, we need to first convert the time to seconds.
5 minutes = 5 min * 60 s/min = 300 s
The distance can be calculated using the formula:
Distance = Speed * Time
Distance = (3.61 m/s) * (300 s)
≈ 1083 m
Therefore, the boat travels approximately 1083 meters in 5 minutes.
d) To calculate the additional force needed to give the boat an initial acceleration of 0.025 m/s², we can use Newton's second law of motion:
Force = Mass * Acceleration
Force = (3000 kg) * (0.025 m/s²)
= 75 N
Therefore, the additional force required is 75 Newtons.