1) A tour boat travels 25 km due east and then 15 km S50°E. Represent these displacements in a vector diagram, then calculate the resultant displacement.
To represent the displacements in a vector diagram, start by drawing a line segment representing 25 km due east. Label it East (E). Then, starting from the endpoint of the first line segment, draw a line segment 15 km long in the direction S50°E. This means the angle between the line segment and the south direction is 50°. Label this line segment S50°E.
To calculate the resultant displacement, we need to find the vector sum of the two displacements. You've correctly drawn the vectors head to tail to form a triangle. Now, you can use the Law of Cosines to find the magnitude of the resultant displacement.
|r|² = 25² + 15² - 2(25)(15)cos130
|r|² = 625 + 225 - 750cos130
|r|² = 850 - 750cos130
|r|² ≈ 850 - 750(-0.4226)
|r|² ≈ 850 + 316.95
|r|² ≈ 1166.95
|r| ≈ √1166.95
|r| ≈ 34.1 km (rounded to one decimal place)
To find the direction of the resultant displacement, you can use the Law of Sines. From your calculations, it seems you used the wrong angle when setting up the equation.
(sin50°/34.1) = (sinC/25)
sinC = (34.1)(sin50°)/25
C ≈ arcsin((34.1)(sin50°)/25)
C ≈ 70.4°
Since the boat traveled south of east, the direction should be measured clockwise from north. Therefore, the resultant displacement is approximately 34.1 km at an angle of 70.4° clockwise from north. You can round the angle to the nearest degree if required.
The correct answer should be 34.1 km S70°E (rounded to one decimal place).
2) Vectors a and b have magnitudes 2 and 3, respectively. If the angle between them is 50°, find the vector "5a - 2b", and state its magnitude and direction.
To find the vector "5a - 2b", you need to subtract vector b from vector 5a.
5a - 2b = 5(2a) - 2(3b) = 10a - 6b
Now, let's find the magnitude of this vector:
|r|² = (10)² + (-6)² - 2(10)(-6)cos50°
|r|² = 100 + 36 + 120cos50°
|r|² ≈ 136 + 120(0.6428)
|r|² ≈ 136 + 77.136
|r| ≈ √213.136
|r| ≈ 14.6 (rounded to one decimal place)
To find the direction of the vector, you can use the Law of Sines. However, it seems you used the wrong angle and magnitudes in your calculation.
(sinθ/3) = (sin50°/14.6)
sinθ ≈ (3)(sin50°)/14.6
θ ≈ arcsin((3)(sin50°)/14.6)
θ ≈ 37.0° (rounded to one decimal place)
Since the vector a was given as the reference, the direction should be measured clockwise from vector a. Therefore, the vector "5a - 2b" has a magnitude of approximately 14.6 and a direction of 37.0° clockwise from vector a.
The correct answer is 14.6, 37° clockwise from vector a (rounded to one decimal place).