Respuesta :
Answer: magnitude of resultant = 2.811km, direction of resultant = 209.6°
Explanation: The vector ( in this case displacement) that lies on the y axis is 3.25km north and 1.50km due south.
Their resultant is gotten below
Magnitude of resultant = 3.25 - 1.50 = 1.75km
Direction of resultant = north (direction of the bigger vector)
2.20km is the only vector acting on the x axis due east, combining this vector with the resultant of the vectors above, we realize that 2.2km west is perpendicular to 1.75km due north. Since 1.75km us due north, it implies that it is the vector on the positive y axis (vy) and 2.20km due west implies that it is the vector on the negative x axis(vx), thus their resultant is gotten using phythagoras theorem.
R = √ vx² + vy²
R = √ 2.20² + 1.75²
R = √ 7.9025
R= 2.1811km.
θ = tan^-1 (vy/vx)
θ = tan ^-1 (1.25/2.20)
θ = tan ^-1 (0.5618)
θ = 29.6°.
The direction of the vector is south west which implies the third quadrant of the trigonometric quadrant which implies 180 + θ
Thus the direction of the vector is 180 + 29.60 = 209.6°
From the picture attached to this answer we can see roughly that the magnitude of resultant is longer than it component and the vector is placed on the 3rd quadrant which verifies our quantitative claim.
The magnitude of the resultant vector can be calculated by the Pythagorean theorem. The magnitude and direction of the resulting vector will be 2.18 km and 29.66°.
The magnitude of resulting vector:
The magnitude of the resultant vector can be calculated by the Pythagorean theorem,
[tex] R = \sqrt { 2.20^2 + 1.75^2}\\\\ R =\sqrt { 7.9025}\\\\ R= 2.1811\rm \ km. [/tex]
The direction of the vector can be calculated by the following equation:
[tex]\Theta = tan^{-1} \dfrac {vy}{vx}[/tex]
Where,
[tex]v_x[/tex] - horizontal distance
[tex]v_y [/tex]- verticle distance
Put the values in the formula,
[tex]\Theta= tan ^-1 \dfrac {1.25}{2.20}\\\\ \Theta = tan ^-1 (0.5618)\\\\ \Theta = 29.66^o [/tex]
Therefore, the magnitude and direction of the resulting vector will be 2.18 km and 29.66°.
Learn more about the magnitude of the resulting vector:
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