A boat sails 5.0(45 degrees W of N). It then hanges direction and sails 7.0km(45 degrees S of E) where does the boat end up with reference to its starting point.?
Please explain steps
answer is 2km(45 degrees S of E)
A = 5.0km[135o]
B = 7.0km[315o]
D = A+B
D=(5.0*cos135+7.0*cos315)+i(5.0*sin135+
7.0*sin315
D = (-3.54+4.95) + i(3.54-4.95)
D = 1.41 - i1.41 = 2km[-45o]=2km[45o S of E].
NOTE: The cosine function was used to
find the hor. components; the sine function was used to find the ver. component.
To determine where the boat ends up with reference to its starting point, we can break down the boat's motion into two components: north-south (N-S) and east-west (E-W).
Step 1: Boat's motion 5.0 km (45 degrees W of N)
First, let's analyze the boat's motion 5.0 km (45 degrees west of north).
To find the N-S and E-W components of the boat's motion, we'll use trigonometry.
The N-S component of the boat's motion can be calculated as follows:
N-S = 5.0 km * sin(45 degrees)
N-S = 5.0 km * 0.707 (rounded to three decimal places)
N-S = 3.54 km (rounded to two decimal places)
Similarly, the E-W component of the boat's motion can be calculated as:
E-W = 5.0 km * cos(45 degrees)
E-W = 5.0 km * 0.707 (rounded to three decimal places)
E-W = 3.54 km (rounded to two decimal places)
Step 2: Boat's motion 7.0 km (45 degrees S of E)
Next, let's analyze the boat's motion 7.0 km (45 degrees south of east).
Again, we'll use trigonometry to find the N-S and E-W components of the boat's motion.
The N-S component of the boat's motion can be calculated as follows:
N-S = 7.0 km * sin(45 degrees)
N-S = 7.0 km * 0.707 (rounded to three decimal places)
N-S = 4.95 km (rounded to two decimal places)
Similarly, the E-W component of the boat's motion can be calculated as:
E-W = 7.0 km * cos(45 degrees)
E-W = 7.0 km * 0.707 (rounded to three decimal places)
E-W = 4.95 km (rounded to two decimal places)
Step 3: Determine the final position
To find the final N-S and E-W position, we need to sum the components separately.
Final N-S position = 3.54 km - 4.95 km = -1.41 km (south)
Final E-W position = 3.54 km + 4.95 km = 8.49 km (east)
Therefore, the boat ends up 2 km (45 degrees south of east) from its starting point.
To find where the boat ends up with reference to its starting point, we can use vector addition.
Step 1: Convert the given distances and angles into their respective vector form.
The first leg of the journey can be represented as a vector pointing 5.0 km in the direction 45 degrees West of North. To convert this to vector form, we use the fact that North is the positive y-direction and East is the positive x-direction in a Cartesian coordinate system.
The vector for the first leg can be represented as:
v1 = 5.0 km * (-cos(45), sin(45))
= 5.0 km * (-0.707, 0.707)
= (-3.535 km, 3.535 km)
The second leg of the journey can be represented as a vector pointing 7.0 km in the direction 45 degrees South of East.
The vector for the second leg can be represented as:
v2 = 7.0 km * (cos(45), -sin(45))
= 7.0 km * (0.707, -0.707)
= (4.949 km, -4.949 km)
Step 2: Add the two vectors to find the resultant vector.
To find the resultant vector, we simply add the two vectors:
resultant = v1 + v2
= (-3.535 km, 3.535 km) + (4.949 km, -4.949 km)
= (1.414 km, -1.414 km)
Step 3: Convert the resultant vector back into angle and distance form.
To convert the resultant vector back into angle and distance form, we first find the magnitude (distance) of the resultant vector:
magnitude = sqrt((1.414 km)^2 + (-1.414 km)^2)
= sqrt(1.998 km^2)
≈ 1.414 km
Next, we calculate the angle of the resultant vector:
angle = tan^(-1)(-1.414 km / 1.414 km)
≈ -45 degrees
Since the angle is negative, we need to add 180 degrees to get the proper direction:
angle = -45 degrees + 180 degrees
= 135 degrees
Therefore, the boat ends up 1.414 km (approximately 2 km) in distance, at an angle of 135 degrees South of East with reference to its starting point.