Question

Select the correct vectors. Bruce's speed when swimming in still water is 5 meters/second. He is swimming in a direction 25° east of north. The current is moving 3.5 meters/second at an angle of 60° west of south. Identify Bruce's vector, the current's vector, and the vector representing Bruce's actual motion. <2.11, 5.28> <-3.03, -1.75> <5.85cos 50.69, <-3.52, -1.5> 5.85sin 50.69> <3.07cos 87.25, <2.93cos 108.26, 3.07 sin 87.25> 2.93sin 108.26> <2.11, 4.53> <3.02cos 121.54, 3.02 sin 121.54>

Answer 1

Bruce's vector is <4.53, 2.11> m/s, the current's vector is <-3.03, -1.75> m/s, and the vector representing Bruce's actual motion is <1.5, 0.36> m/s.

Step-by-step explanation:

To determine Bruce's vector, we need to break down his speed into its horizontal and vertical components. Bruce's speed is 5 m/s and he is swimming in a direction 25° east of north. Using trigonometry, we can find that the horizontal component, or the vector in the east-west direction, is 5 * cos(25°) = 4.53 m/s, and the vertical component, or the vector in the north-south direction, is 5 * sin(25°) = 2.11 m/s.

The current's vector is given as <-3.03, -1.75> m/s, which represents the west-east and south-north components respectively. The vector representing Bruce's actual motion is obtained by adding Bruce's and the current's vectors componentwise. So, the resulting vector is <4.53 + (-3.03), 2.11 + (-1.75)> = <1.5, 0.36> m/s.

Answer 2

Bruce's vector representing his speed in still water is <2.11, 4.53>. The current's vector representing its velocity is <-3.02cos 121.54, 3.02 sin 121.54>. Bruce's vector representing his actual motion is <5.85cos 50.69, -3.52, -1.5> or approximately <5.85, -4.02, -1.5>.

Step-by-step explanation:

The vector representing Bruce's speed in still water is <2.93cos 108.26, 3.07 sin 87.25>. This vector can be broken down into its x and y components, giving us approximately <2.11, 4.53>. The vector representing the current's velocity is <-3.02cos 121.54, 3.02 sin 121.54>.

Lastly, the vector representing Bruce's actual motion is the sum of his speed and the current's velocity, giving us <5.85cos 50.69, -3.52, -1.5> or approximately <5.85, -4.02, -1.5>.