Question

1. A ship sails at a constant speed of 20 km/hr. The ship leaves port and sails 20 degrees north of east for 2 hours. The ship then changes direction and sails 10 degrees west of north for 1 hours. At the end of this trip how far is the ship from the starting port ?

Answer 1

The ship sails at 20 km / h.
 So
 20 km / h x 2 h = 40 km
 The boat moves 40 km to the northeast.
 Then, in the same way:
 20 km / h x 1 h = 20 km.
 The ship moves 20 km northwest.
 We solve this problem using vectors. In this case, we must perform the sum of two vectors a and b
 a) magnitude = 40 km and direction 20 degrees east
 b) magnitude = 20 km and direction 10 degrees northwest
 In Cartesian coordinates, these vectors are written as:
 a) 40sin (20º) i + 40cos (20º) j = 13,68i + 37,59j
 b) -20sin (10th) i + 20cos (10th) j = -3,473i + 19,70j
 The displacement vector of the ship would be the sum of a + b
 a + b = 10,207i + 57,29j Finally, the magnitude of the a + b vector will tell us how far the ship is
 √(10.21² + 57.29²) = 58.92km

Answer 2

The ship is 41.23 km away from the starting port.

Calculus

Given that a ship sails at a constant speed of 20 km/hr, and the ship leaves port and sails 20 degrees north of east for 2 hours, and the ship then changes direction and sails 10 degrees west of north for 1 hours, to determine , at the end of this trip, how far is the ship from the starting port, the following calculation must be made, applying the Pythagorean theorem:

• 40^2 + 10^2 = X^2
• 1600 + 100 = X^2
• √1700 = X
• 41.23 = X

Therefore, the ship is 41.23 km away from the starting port.

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