Person A walks 6.0 m at an angle of 30 north of east. Person B walks 7.0 m at an angle of 42 west of north. How does the component of the displacement of person A along the north direction compare with the component of the displacement of person B along the north direction?
A.
It is larger.
B.
It is smaller.
C.
They are equal.
D.
There is not sufficient information to tell.
A = 6m[30o[
Y = 6*sin30 = 3m = North Component.
B = 7m[132o], CCW.
Y = 7*sin132 = 5,20 m. = North component.
The component along the north direction
of person A is smaller.
To determine the component of the displacement along the north direction for person A and person B, we need to find the northward component of their respective displacements.
For person A:
The displacement of person A can be split into two components: one along the east direction and one along the north direction. Since person A walks at an angle of 30° north of east, we can use trigonometry to find the northward component.
Using the cosine function:
cos(30°) = adjacent/hypotenuse
cos(30°) = north-component of displacement/6.0 m
Solving for the north-component of displacement:
north-component of displacement = cos(30°) * 6.0 m
north-component of displacement = (0.866) * 6.0 m ≈ 5.196 m
For person B:
The displacement of person B can also be split into two components: one along the west direction and one along the north direction. Since person B walks at an angle of 42° west of north, we can use trigonometry to find the northward component.
Using the sine function:
sin(42°) = opposite/hypotenuse
sin(42°) = north-component of displacement/7.0 m
Solving for the north-component of displacement:
north-component of displacement = sin(42°) * 7.0 m
north-component of displacement = (0.669) * 7.0 m ≈ 4.683 m
Comparing the northward components of the displacements:
5.196 m (person A) > 4.683 m (person B)
Therefore, the component of the displacement of person A along the north direction is larger than the component of the displacement of person B along the north direction.
Answer: A. It is larger.
To determine how the component of person A's displacement along the north direction compares with the component of person B's displacement along the north direction, we need to calculate these components separately.
Component of person A's displacement along the north direction:
Since person A walks at an angle of 30° north of east, we can use trigonometry to find the north component.
Draw a right-angled triangle where the hypotenuse represents the total displacement of person A, the adjacent side represents the east component, and the opposite side represents the north component.
We can use the equation cosine(theta) = adjacent/hypotenuse to find the east component.
cos(30°) = east component / 6.0 m
east component = 6.0 m * cos(30°)
east component ≈ 5.20 m
To find the north component, we can use the equation sine(theta) = opposite/hypotenuse.
sine(30°) = north component / 6.0 m
north component = 6.0 m * sine(30°)
north component ≈ 3.00 m
Component of person B's displacement along the north direction:
Since person B walks at an angle of 42° west of north, we can again use trigonometry to find the north component.
Draw a right-angled triangle where the hypotenuse represents the total displacement of person B, the adjacent side represents the north component, and the opposite side represents the west component.
We can use the equation sine(theta) = opposite/hypotenuse to find the north component.
sine(42°) = north component / 7.0 m
north component = 7.0 m * sine(42°)
north component ≈ 4.71 m
Now we can compare the north components of person A and person B:
Person A's north component ≈ 3.00 m
Person B's north component ≈ 4.71 m
Since 4.71 m is greater than 3.00 m, the component of the displacement of person B along the north direction is larger.
Therefore, the answer is A. It is larger.