At the beginning of a race, a yacht crosses the start line and sails in
westerly direction at a speed of 4 m s-1. The current of the sea is flowing at
2 m s-1 towards the north-east. Calculate how far [8 marks] and in what
direction (relative to west) [8 marks] the yacht has traveled after 10
minutes from its crossing the start line.
The answer is
(-4,0) + (1.4,1.4) = (-2.6,1.4)
= 2.96 m/s at W28.3°N
After 10 minutes (600 seconds) the distance traveled is 1772m
Can you show me how to work this out. thanks!
4m/s @ 180 Deg + 2m/s @ 45 Deg.
X = 4*cos180 + 2*cos45 = -2.586 m/s.
Y = 0 + 2*sin45 = 1.414 m/s.
tanAr=Y/X = 1.414 / -2.586 = -.054687,
Ar = -28.7 Deg. = Reference angle.
A=-28.7 + 180=151.3 Deg.,CCW= W28.7Deg
N.
V = X/cosA = -2.586 / cos151.3 = 2.95 m/s @ W28.7Deg.N.
d = 2.95m/s * 600s = 1770 m.
Sure, I can help you with that!
To calculate how far the yacht has traveled after 10 minutes, we need to find the resultant velocity vector. We can do this by adding the velocity of the yacht and the velocity of the current.
The velocity of the yacht is given as 4 m/s in the westerly direction. So its velocity vector can be represented as (4, 0) m/s.
The velocity of the current is given as 2 m/s towards the northeast. To represent this vector, we need to break it down into its northward and eastward components. Since the current is flowing at a 45-degree angle between north and east, the northward component would be 2*sin(45) = 2*(1/sqrt(2)) = 1.4 m/s, and the eastward component would also be 1.4 m/s. So the velocity vector of the current can be represented as (1.4, 1.4) m/s.
Now we can add the two velocity vectors:
(4, 0) + (1.4, 1.4) = (4+1.4, 0+1.4) = (5.4, 1.4) m/s
So after 10 minutes, the yacht has traveled a distance of sqrt((5.4^2) + (1.4^2)) = sqrt(29.16 + 1.96) = sqrt(31.12) ≈ 5.57 m.
To find the direction relative to the west, we can use the trigonometric function arctan. The angle θ can be found using:
θ = arctan(1.4/5.4) ≈ 14.2 degrees north of west.
So the yacht has traveled approximately 5.57 m at an angle of 14.2 degrees north of west after 10 minutes from crossing the start line.
To solve this problem, we can break it down into two components: the westward component and the northward component of the yacht's motion.
First, let's calculate the westward distance traveled:
Speed = 4 m/s
Time = 10 minutes = 10 x 60 = 600 seconds
Westward distance = Speed x Time = 4 m/s x 600 s = 2400 m west
Next, let's calculate the northward distance traveled:
Speed of the current = 2 m/s
Time = 10 minutes = 10 x 60 = 600 seconds
Northward distance = Speed x Time = 2 m/s x 600 s = 1200 m north
We have the westward distance of 2400 m and the northward distance of 1200 m. To find the total distance traveled and the direction, we can use vector addition.
The position vector of the yacht's motion is given by:
(-2400, 0) + (0, 1200) = (-2400, 1200)
To find the resultant vector, take the square root of the sum of the squares of the components:
Resultant vector = √((-2400)^2 + (1200)^2) = √(5760000 + 1440000) = √7200000 = 2682.31 m
The direction is given by the arctan of the northward component over the westward component:
Direction = arctan(1200/2400) = arctan(0.5) = 26.6°
Therefore, the yacht has traveled a total distance of 2682.31 m at a direction of 26.6° relative to the west after 10 minutes.
Sure! To calculate how far the yacht has traveled after 10 minutes, we need to consider both the speed of the yacht and the current of the sea.
First, let's calculate the distance traveled by the yacht in 10 minutes. Given that the yacht's speed is 4 m/s, we can use the formula:
Distance = Speed × Time
Distance = 4 m/s × 600 s = 2400 m
So, the yacht has traveled a distance of 2400 meters in 10 minutes.
Next, let's calculate the displacement of the yacht, taking into account the current of the sea. We will use vector addition to calculate this.
The yacht is traveling westward at 4 m/s, which can be represented as (-4, 0) in terms of x and y components.
The current of the sea is flowing at 2 m/s towards the northeast, which can be represented as (+1.4, +1.4) in terms of x and y components.
Now, we add these two vectors together:
(-4, 0) + (1.4, 1.4) = (-2.6, 1.4)
So, the displacement of the yacht after 10 minutes is (-2.6, 1.4) in terms of x and y components.
To find the magnitude and direction of this displacement, we can use the Pythagorean theorem and trigonometry.
The magnitude of the displacement can be found using the formula:
Magnitude = √(x^2 + y^2)
Magnitude = √((-2.6)^2 + (1.4)^2) = √(6.76 + 1.96) = √8.72 ≈ 2.96 m/s
The direction of the displacement relative to west can be found using the formula:
Angle = atan(y / x)
Angle = atan(1.4 / -2.6) ≈ atan(-0.538) ≈ -28.3°
Since the yacht is traveling westward, we would subtract 28.3° from 180° to get the final direction:
Angle = 180° - 28.3° ≈ 151.7°
So, the yacht has traveled approximately 2.96 m/s at a direction of W28.3°N.
Therefore, after 10 minutes from its crossing the start line, the yacht has traveled approximately 1772 meters and it is moving at 2.96 m/s in the direction of W28.3°N.