Question

A boomerang is thrown with the following
displacements: 4.5 m, E, 7.5 m, 30 degrees N of
W, and 3.0 m at 10 degrees S of W. What is the
resultant displacement?

Answer 1

Let us denote the net three consecutive displacements by the vectors A,B,C

while R represents the resultant displacement.

Displacement A has a magnitude of 4.5m and it's components are;

$A_x = Acos(0) = 4.5m × 1 = 4.5m\\ A_y = Asin(0) = 4.5m × 0 = 0m\\$

The second displacement B has a magnitude of 7.5m and it's components are;

$B_x = Bcos(150) = 7.5m × (0.866) = -6.495m\\ B_y = Bsin(150) = 7.5 × (0.5) = 3.75m\\$

Finally, displacement C whose magnitude is 3.0m has the components;

$C_x = Ccos(190) = 3m × (-0.9848) = -2.954m\\ C_y = Csin(190) = 3m × (-0.1736) = -0.521m\\$

The resultant displacement R = A+B+C has components;

$R_x = A_x+B_x+C_x = 4.5m-6.495m-2.954m\\ R_x=-4.949m\\ R_y=A_y+B_y+C_y=0m+3.75m-0.521m\\ R_y=3.229m$

In unit vector form, we can write the total displacement as,

$R= (-4.949i+3.229j)m\\$

The magnitude of R is therefore,

$\sqrt{(-4.949m)^2+(3.229m)^2} = 5.91m$

It's direction is found by calculating the angle it makes with the x-axis.

$tan\theta=\frac{R_y}{R_x}=\frac{3.229}{-4.949}= -33.1°\\ i.e 56.88° \textsf{W of N}\\ \textsf{The resultant displacement is therefore 5.91m,56.88° W of N}$