A ship leaves Tema harbor at 10:00am one morning. The captain steers a course bearing 136° at a speed of 20km/h through the water. There is a current of 5km/h in a direction of 046°. Find
i. the direction in which the boat travels.
ii. the distance from Tema at 10:00am.
b. Let â be a unit vector making an angle α(0<α<π/2), measured in anti-clockwise direction, with positive x axis and b ̂ be unit vector making an angle β(0<β<□(π ̅/2)), measured in the anti-clockwise direction with positive x axis.
i. What is the angle between a ̂ andb ̂? Hence write down an expression for a ̂ . b ̂.
ii. Write down the components of a ̂ andb ̂. Hence write down another expression for a ̂ . b ̂.
iii. Deduce the result cos(α-β)= cosαcosβ+sinαsinβ.
You steer a heading, not a bearing. There is a conspiracy of inland math teachers trying to drive navigators crazy.
North component of speed through water
= 20 cos 136 = -14.4 km/hr
East component of speed through water
= 20 sin 136 = +13.9 km/hr
North component of current
= 5 cos 46 = 3.47 km/hr
East component of current
= 5 sin 46 = 3.60 km/hr
so we are headed
North at -10.9 km/hr and East at +17.5 km/hr
Theta is angle from vertical axis
tan Theta = 17.5/-10.9
Theta = -58 degrees or 58 from south which is 122 clockwise from north
speed = sqrt (10.9^2 + 17.5^2)
distance = speed * time
ii is a typo. You left at 10 am so you are still there.
in general
North distance = -10.6 t
East distance = 3.6 t
I need further explanation on this because I was given the same question:
1). A ship leaves Lagos Barbour at 10am one morning. The captain steers a course bearing 136° at a speed of 20km/hr through the water. There is a current of 5km/hr in the direction 046°. Find i) the direction in which the boat steer. ii) its distance from Lagos at 1.00 pm.
To solve this problem, we will use vector addition and trigonometry.
i. To find the direction in which the boat travels, we need to determine the resultant velocity vector by considering both the boat's velocity and the current's velocity.
Let's define vectors:
- Boat's velocity vector: B = 20 km/h at a heading of 136°
- Current's velocity vector: C = 5 km/h at a heading of 046°
To find the resultant velocity vector, we add both vectors: R = B + C
To add vectors, we need to break them down into their x and y components. Using trigonometry, we can calculate the components of each vector:
- For B:
- Bx = magnitude of B * cos(bearing of B) = 20 km/h * cos(136°)
- By = magnitude of B * sin(bearing of B) = 20 km/h * sin(136°)
- For C:
- Cx = magnitude of C * cos(bearing of C) = 5 km/h * cos(46°)
- Cy = magnitude of C * sin(bearing of C) = 5 km/h * sin(46°)
Next, we add the x-components and the y-components separately:
- Rx = Bx + Cx
- Ry = By + Cy
The direction in which the boat travels can be found using the inverse tangent function:
- Direction = arctan(Ry / Rx)
ii. To find the distance from Tema at 10:00 am, we need to calculate the magnitude of the resultant velocity vector.
The magnitude of the resultant velocity vector is given by the Pythagorean theorem:
- Magnitude = sqrt(Rx^2 + Ry^2)
b. Now let's move on to the second part:
i. The angle between two unit vectors a ̂ and b ̂ can be found using the dot product formula:
- Angle between a ̂ and b ̂ = arccos(a ̂ . b ̂)
The expression for the dot product can be written as:
- a ̂ . b ̂ = a1 * b1 + a2 * b2
ii. The components of vectors a ̂ and b ̂ can be found using trigonometry and the given angles α and β:
For a ̂:
- ax = cos(α)
- ay = sin(α)
For b ̂:
- bx = cos(β)
- by = sin(β)
Using the components, the expression for the dot product becomes:
- a ̂ . b ̂ = (ax * bx) + (ay * by)
iii. To deduce the result cos(α-β) = cos(α) * cos(β) + sin(α) * sin(β), we need to use the angle difference trigonometric identity.
The angle difference identity states that:
- cos(α-β) = cos(α) * cos(β) + sin(α) * sin(β)
Therefore, the result is deduced.