adaobi walks 2km due west and 2km due south calculate the displacement of the girl
X = -2 km.
Y = -2 km.
D^2 = X^2 + Y^2,
D^2 = (-2)^2 + (-2)^2 = 8,
D = 2.83 km.
tanA = Y/X = -2 / -2 = 1.
Ar = 45 Deg. = Reference angle.
A = 45 + 180 = 225 Deg., Q3.
D = 2.83 km @ 225 Deg.
thanks
X=v×t
C^2=(2)^2+(2)^2
C^2=4+4
C^2=8
Tan=opp÷adj=2÷2=1
Tan-11=45
X=2Square root 2
=45dues
To calculate the displacement of Adaobi, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
In this case, Adaobi walks 2 km due west (left) and 2 km due south (down). This forms a right-angled triangle with sides of length 2 km and 2 km.
To find the displacement, we need to find the length of the diagonal, which is the hypotenuse of the triangle. We can use the formula:
displacement = √(side1^2 + side2^2)
In this case, side1 = 2 km and side2 = 2 km. Substituting these values into the formula:
displacement = √(2 km)^2 + (2 km)^2
Simplifying:
displacement = √(4 km^2 + 4 km^2)
displacement = √8 km^2
Taking the square root:
displacement ≈ 2.83 km
Therefore, the displacement of Adaobi is approximately 2.83 km.