solve the mathematics question.Two ships leaves port at the same time,one steaming 5km/hr on a bearing N46E and the other at 9km/hr on a bearing S53E.How far apart are the ships after 2hours.
Draw the diagram, and note the angle between the two directions.
distance = speed * time
then use the law of cosines to find the desired distance
the angle between the two headings is ... 180 - 46 - 53 = 81
the distances traveled after 2 hr are
... 2 h * 5 kph = 10 km
... 2 hr * 9 kph = 18 km
use the law of cosines to find the distance (d) between the ships
d = √{10^2 + 18^2 - [2 * 10 * 18 * cos(81º)]}
To solve this question, we can use the concept of vector addition.
First, let's break down the velocities of both ships into their respective North (N) and East (E) components.
For Ship 1:
Velocity = 5 km/hr
Bearing = N46E
To find the North and East components, we can use trigonometry. Using the given bearing, we can determine the angles involved:
Angle N46E = 90° - 46° = 44°
To find the North component:
North Component = Velocity * sin(Angle N46E)
= 5 km/hr * sin(44°)
≈ 3.54 km/hr
To find the East component:
East Component = Velocity * cos(Angle N46E)
= 5 km/hr * cos(44°)
≈ 3.82 km/hr
Similarly, for Ship 2:
Velocity = 9 km/hr
Bearing = S53E
Angle S53E = 90° + 53° = 143° (since we need to consider the direction south of east)
To find the North component:
North Component = Velocity * sin(Angle S53E)
= 9 km/hr * sin(143°)
≈ -7.72 km/hr (negative sign indicates South direction)
To find the East component:
East Component = Velocity * cos(Angle S53E)
= 9 km/hr * cos(143°)
≈ -4.99 km/hr (negative sign indicates West direction)
Now, we can calculate the displacement between the ships after 2 hours. Displacement is calculated separately for both the North and East directions and then combined using vector addition.
For the North direction:
Displacement North = (North Component of Ship 1 * time) + (North Component of Ship 2 * time)
= (3.54 km/hr * 2 hr) + (-7.72 km/hr * 2 hr)
= 7.08 km - 15.44 km
≈ -8.36 km
For the East direction:
Displacement East = (East Component of Ship 1 * time) + (East Component of Ship 2 * time)
= (3.82 km/hr * 2 hr) + (-4.99 km/hr * 2 hr)
= 7.64 km - 9.98 km
≈ -2.34 km
Finally, to find the distance (magnitude) between the ships after 2 hours, we can use the Pythagorean theorem:
Distance between ships = √((Displacement North)^2 + (Displacement East)^2)
= √((-8.36 km)^2 + (-2.34 km)^2)
≈ √(69.92 km^2 + 5.48 km^2)
≈ √75.4 km^2
≈ 8.68 km
Therefore, the ships are approximately 8.68 km apart after 2 hours.