A Studet walked 4km due east and the 15km east of north in direaction of 37° north of east.finally he walked due north for 3km.how much is displacement.
starting at (0,0) the student walked to
(x,y) = (4+15sin37° , 15cos37°+3)
the displacement is √(x^2+y^2)
in the direction NθE where tanθ = x/y
Teachers
To teach the students in different ways.
To find the displacement of the student, we can break down the given directions into their respective components and then sum them up.
Let's analyze each part:
1. The student walked 4 km due east.
The displacement in the east direction is +4 km, as the student is moving directly towards the east.
2. The student then walked 15 km east of north.
The direction is given as 37° north of east, which means the student is moving at an angle 37° above the east direction. To find the eastward and northward components of this movement, we can refer to the triangle formed by the displacement vector, the eastward component, and the northward component.
Using trigonometry, we can calculate the eastward component:
eastward component = magnitude of movement * cos(angle)
= 15 km * cos(37°)
≈ 11.94 km (rounded to two decimal places)
And the northward component:
northward component = magnitude of movement * sin(angle)
= 15 km * sin(37°)
≈ 9.05 km (rounded to two decimal places)
3. Finally, the student walked due north for 3 km.
The displacement in the north direction is +3 km, as the student is moving directly towards the north.
Now, let's sum up the components:
Eastward displacement: +4 km (from step 1)
Northward displacement: +9.05 km (from step 2)
Vertical displacement: +3 km (from step 3)
Using the Pythagorean theorem, we can find the resultant displacement (R):
R = √(eastward displacement^2 + northward displacement^2 + vertical displacement^2)
= √((4 km)^2 + (9.05 km)^2 + (3 km)^2)
≈ 10.91 km (rounded to two decimal places)
Therefore, the displacement of the student is approximately 10.91 km.