Asked by girasa
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.
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.
Answers
Answered by
Damon
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
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
Answered by
Daviski
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.
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.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.