Question

Find the x- and y-components of the vector a⃗ = (19 m/s2 , 31 ∘ left of −y-axis).

Answer 1

The value of the acceleration given is,

$a=19ms^{-2}$

The angle of the acceleration with the negative of the y-axis in the third quadrant is,

$\theta=31^{\circ}$

Solution:

The diagrammatic representation of the given system is,

The angle of the acceleration with the positive of the x-axis is,

$\begin{gathered} \theta_1=31^{\circ}+90^{\circ} \\ \theta_1=121^{\circ} \end{gathered}$

Thus, the component of the acceleration along the positive of the x-axis is,

$\begin{gathered} a_x=a\cos (\theta_1) \\ a_x=19\cos (121^{\circ}) \\ a_x=-9.8ms^{-2} \end{gathered}$

The angle of the acceleration with the positive of y-axis is,

$\begin{gathered} \theta_2=(90^{\circ}-31^{\circ})+90^{\circ} \\ \theta_2=149^{^{\circ}} \end{gathered}$

The y-component of the acceleration is,

$\begin{gathered} a_y=a\cos (\theta_2) \\ a_y=19\times\cos (149^{\circ}) \\ a_y=-16.29ms^{-2} \end{gathered}$

Thus, the x and y-component of the acceleration are -9.8,-16.29.