a yacht is moving at 10km/h in a south easterly direction and encounters a 3km/h current from the north. find the actual speed and direction of the yacht
Well, it seems like the yacht is dealing with some water shenanigans. Let's break this down, shall we?
The yacht is moving at 10 km/h in a south-easterly direction. Meanwhile, it's also encountering a 3 km/h current coming from the north. Oh, those sneaky currents!
To find the actual speed and direction of the yacht, we need to consider these two motions together. So, we can imagine them as two vectors: one representing the yacht's velocity and the other representing the current's velocity.
Now, let's do a little math circus. If we add these vectors, the actual speed of the yacht would be the magnitude of the resultant vector, and the direction would be the angle it makes with the initial direction.
Using some trigonometry and vector addition, the actual speed would be about 10.44 km/h (approximately) and the direction would be somewhere between south-east and east-south-east. So, we can say it's wobbling along in that general direction.
So, to put it simply, the yacht's actual speed is around 10.44 km/h in a wobbly south-easterly, east-south-easterly direction. Keep navigating those waves, sailor!
To find the actual speed and direction of the yacht, we can use vector addition.
Let's break down the given information into the yacht's velocity and the current's velocity.
1. Yacht's velocity:
The yacht is moving at 10 km/h in a south-easterly direction. This can be broken down into two components: one along the south direction and another along the east direction.
- South velocity: The south direction is directly opposite to the north direction. Since the yacht is moving south, the south velocity component is positive.
- East velocity: The east direction is perpendicular to the south direction. Since the yacht is moving in the easterly direction, the east velocity component is also positive.
2. Current's velocity:
The current is moving at 3 km/h from the north direction. Since it is a current coming from the north, the north velocity component is positive.
Now, let's calculate the actual velocity of the yacht by adding the yacht's velocity and the current's velocity.
- South velocity of the yacht = +10 km/h
- East velocity of the yacht = +10 km/h
- North velocity of the current = +3 km/h
To find the actual velocity, we can add the north and south velocities and the east velocity components.
Actual velocity = (East velocity of the yacht + South velocity of the yacht) - North velocity of the current
= (+10 km/h + 10 km/h) - (+3 km/h)
= 20 km/h - 3 km/h
= 17 km/h
The actual speed of the yacht is 17 km/h.
To find the direction, we can use trigonometry. The south and east components of the yacht's velocity form a right triangle with the actual velocity as the hypotenuse.
Using the Pythagorean theorem, we can find the angle of the hypotenuse (actual velocity) with respect to the east direction.
tan(θ) = (South velocity of the yacht)/(East velocity of the yacht)
θ = tan^(-1)((South velocity of the yacht)/(East velocity of the yacht))
θ = tan^(-1)(10/10)
θ = 45 degrees
Therefore, the actual direction of the yacht is 45 degrees south of east.
To find the actual speed and direction of the yacht, we can use vector addition.
Step 1: Draw a diagram to represent the situation. Draw a vector to represent the speed and direction of the yacht and another vector to represent the speed and direction of the current.
Step 2: Determine the x and y components of each vector.
For the yacht's velocity:
- The south easterly direction can be broken down into two components: east and south.
- The magnitude of the yacht's velocity is 10 km/h.
- The east component of the yacht's velocity is 10 km/h multiplied by the cosine of the angle between the yacht's velocity and the east direction.
- The south component of the yacht's velocity is 10 km/h multiplied by the sine of the angle between the yacht's velocity and the south direction.
For the current's velocity:
- The north direction can be broken down into two components: north and east.
- The magnitude of the current's velocity is 3 km/h.
- The north component of the current's velocity is 3 km/h multiplied by the cosine of the angle between the current's velocity and the north direction.
- The east component of the current's velocity is 3 km/h multiplied by the sine of the angle between the current's velocity and the east direction.
Step 3: Add the x and y components of the yacht's and the current's velocity to get the resulting velocity vector.
Step 4: Calculate the magnitude and direction of the resulting velocity vector.
Let's perform these steps:
- For the yacht's velocity:
- The east component is 10 km/h * cos(45°) = 7.07 km/h.
- The south component is 10 km/h * sin(45°) = 7.07 km/h.
- For the current's velocity:
- The north component is 3 km/h * cos(90°) = 0 km/h.
- The east component is 3 km/h * sin(90°) = 3 km/h.
- Adding the x and y components:
- The east component of the resulting velocity is 7.07 km/h + 3 km/h = 10.07 km/h.
- The south component of the resulting velocity is 7.07 km/h + 0 km/h = 7.07 km/h.
- Calculate the magnitude and direction of the resulting velocity:
- The magnitude of the resulting velocity is the square root of (10.07^2 + 7.07^2) km/h, which is approximately 12.82 km/h.
- The direction of the resulting velocity can be found by taking the inverse tangent of the south component divided by the east component, which is atan(7.07/10.07) = 35.34°.
Therefore, the actual speed of the yacht is approximately 12.82 km/h, and its direction is south-east at an angle of 35.34° from the east direction.
I made a sketch and it looks like a simple cosine law problem
d^2 = 10^2 + 3^2 - 2(10)(3)cos135
= 151.426...
d = 12.3 km/hr
then to find the small acute angle Ø
sinØ/3 = sin 135°/12.3..
sinØ = .1723..
Ø = 9.927°
direction is S (45-9.927..)° E
= S 35° E to the nearest degree