Question

A particle undergoes three consecutive displacements. The first is to the east and has amagnitude of 25.0 meters. The second is due north and has a magnitude of 42.0 meters. The thirddisplacement is unknown but the resultant displacement is 38.0 meters [E 30.0° N].Make a vector diagram, to scale, using a ruler and a protractor, in order to determine themagnitude and direction of the third displacement.​

Answer 1

The magnitude of the third vector is 10.9 m and the direction of the third vector is 56.6⁰.

How to calculate the magnitude and direction of third displacement?

The sum of the vectors in x - direction;

25 + Ax = 38 cos(60)

25 + Ax = 19

Ax = 19 - 25

Ax = -6

The sum of the vectors in y - direction;

42 + Ay = 38 sin(60)

42 + Ay = 32.9

Ay = 32.9 - 42

Ay = -9.1

The magnitude of the third vector is calculated as;

A = √ (Ax² + Ay²)

A = √ [(-6)² + (-9.1)²]

A = 10.9 m

The direction of the third vector is calculated as;

θ = arc tan (Ay/Ax)

θ = arc tan (9.1/6)

θ = 56.6⁰