Vectors: A boat moves through the ocean on a foggy day. The boat is pointing North and can typically move at a speed of 12.8 m/s in calm waters. The boat travels through water where there is a 5.9 m/s current flowing to the East. What is the speed of the boat relative to a stationary observer on land? __m/s Part 2 The boat turns to face directly West. Now how fast is the boat moving relative to land? __m/s

1.) 14.09 m/s

2.) 6.9 m/s

Step-by-step explanation:

Given that the boat is pointing North and can typically move at a speed of 12.8 m/s in calm waters. The boat travels through water where there is a 5.9 m/s current flowing to the East.

The speed of the boat relative to a stationary observer on land will be achieved by using pythagorean theorem

V^2 = 12.8^2 + 5.9^2

V^2 = 198.65

Square root both sides

V = 14.09 m/s

2.) If the boat turns to face directly West.

Since the boat can move at speed 12.8 m/s in a calm water, the speed of the boat if it face west will be

12.8 - 5.9 = 6.9 m/s

Therefore, the boat is moving relative to land at 6.9m/s

the distance from any point on the parabola to the focus is the exact same as the distance from that point to the directrix.

if you could provide answer choices that might allow me to answer the question easier if my above answer isn't an option! (:

an

y = 6x -3y -12

y +3y +12 = 6x

=> . x = { 2y / 3 +2 } this is equation of line in terms of x

=> . if line meets the x-axis the co-ordinates of meeting point would be ( x , 0 )

=> . x = 2 ( 0 ) / 3 + 2

=> . x = 2

so, the point is ( 2 , 0 )