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?

Question

Mathematics

Keith Gaines

2 September, 16:03

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?

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Answers (1)

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Ahmed · 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