A ship,A moves due north at a speed of 30km/h while a second ship B moves at a speed of 20km/h on a bearing of 300' determine
(a) the magnitude and direction of the velocity of B relative to A.
(B) the shortest distance of approach of the two ship,if ship B is 10km due east of A.?
You do not move on a "bearing" of 300 deg.
You mean a "heading" of 300 deg.
A bearing is a direction from you to a target.
Mathematicians drive me nuts.
Anyway:
a heading of 300 deg is 60 deg West of North
so
A is 30 km/h at 0 deg
and
B is 20 km /h at 60 deg left of that
so boat B is going 20 cos 60 North and 20 sin 60 west. That is 10 North and 17.3 km/h West.
(a) asks for B - A
north = 10 - 30 = -20 km/h or falling behind at 0 km/h
west = 17.3 - 0 = 17.3 km/h to the left (West, port side of A)
magnitude = sqrt (400 +17.3)^2
tangent of direction West of South = 17.3/ 20
typo
sorry sqrt (400 + 17.3^2)
B-A distance N = 0 - 20 t
B-A distance west = 17.3 t - 10
so
r = [400 t^2 + (17.3 t - 10)^2)]^.5
r = [400 t^2 + (299 t^ 2 - 346 t +100) ]^.5
r = [101 t^2 - 346 t + 100 ]^.5
dr/dt = 0 at desired time
dr/dt = 0 =.5 * (202 t -346) / [101 t^2 - 346 t + 100 ]^.5
so when t = 346/202
then go back and get r (after checking my arithmetic carefully)
obviously wrong
r = [400 t^2 + (299 t^ 2 - 346 t +100) ]^.5
r = [699 t^2 - 346 t + 100 ]^.5
dr/dt = 0 at desired time
dr/dt = 0 =.5 * (1400 t -346) / [101 t^2 - 346 t + 100 ]^.5
so when t = 1400/202 about 7 hours
then go back and get r (after checking my arithmetic carefully)
To determine the magnitude and direction of the velocity of B relative to A, we can use vector addition.
(a) Velocity is a vector quantity that has magnitude and direction. We can break down the velocity of B into its components relative to A.
First, let's find the horizontal and vertical components of B's velocity relative to A.
The horizontal component can be found using trigonometry. The bearing of 300' means that B is moving 60 degrees clockwise from the north direction. We can find the horizontal component by using cosine:
Horizontal component = B's speed * cos(60°)
Horizontal component = 20 km/h * cos(60°)
Horizontal component = 10 km/h
Since B is moving due east, the horizontal component represents the velocity in the eastward direction.
The vertical component can be found using the same trigonometry approach. B's velocity relative to A is due north, so the vertical component represents the velocity in the northward direction.
Vertical component = B's speed * sin(60°)
Vertical component = 20 km/h * sin(60°)
Vertical component = 17.32 km/h
Now we can use Pythagoras' theorem to find the magnitude of B's velocity relative to A:
Magnitude = sqrt((horizontal component)^2 + (vertical component)^2)
Magnitude = sqrt((10 km/h)^2 + (17.32 km/h)^2)
Magnitude = sqrt(100 km^2/h^2 + 300 km^2/h^2)
Magnitude = sqrt(400 km^2/h^2)
Magnitude = 20 km/h
So the magnitude of the velocity of B relative to A is 20 km/h.
To find the direction, we can use inverse trigonometry. The direction is the angle that the velocity vector of B relative to A makes with the north direction.
Direction = atan2(vertical component, horizontal component)
Direction = atan2(17.32 km/h, 10 km/h)
Direction ≈ 59.04°
Therefore, the magnitude of the velocity of B relative to A is 20 km/h, and the direction is approximately 59.04° from the north direction.
(B) To determine the shortest distance of approach between the two ships, we can use the concept of relative motion.
First, let's determine how long it would take for ship A to cover the horizontal distance of 10 km (since ship B is 10 km due east of A).
Time = distance / speed
Time = 10 km / 30 km/h
Time = 1/3 hour
Now, let's consider ship B's motion during the same amount of time. Ship B is moving at a speed of 20 km/h, and it would take 1/3 hour for ship A to cover the 10 km distance.
Distance traveled by ship B during 1/3 hour = speed * time
Distance traveled by ship B = 20 km/h * 1/3 hour
Distance traveled by ship B = 6.67 km
Therefore, ship B would have traveled a horizontal distance of 6.67 km during the 1/3 hour when ship A covers the 10 km distance due north.
To find the shortest distance of approach, we need to find the perpendicular distance between the positions of ship A and ship B at the end of the 1/3 hour. This can be thought of as a right-angled triangle, with the horizontal distance traveled by ship B (6.67 km) as the base and the perpendicular distance as the height.
By using Pythagoras' theorem, we can find the shortest distance of approach:
Shortest distance of approach = sqrt((horizontal distance)^2 + (perpendicular distance)^2)
Shortest distance of approach = sqrt((6.67 km)^2 + (10 km)^2)
Shortest distance of approach = sqrt(44.44 km^2 + 100 km^2)
Shortest distance of approach = sqrt(144.44 km^2)
Shortest distance of approach ≈ 12 km
Therefore, the shortest distance of approach between the two ships, given that ship B is 10 km due east of A, is approximately 12 km.