91Ó°ÊÓ

Particle A moves along the${}^{\mathrm{y}=30\mathrm{m}}$with constant velocity${}^{\mathrm{v}=3\mathrm{m}/s}$parallel to x axis.

Particle B moves with${}^{v=0\mathrm{m}/s}$and constant acceleration ${}^{\stackrel{â‡\P Ä}{\mathrm{a}}=0.40m/s^2}$

This problem is based on kinematic equations that describe the motion of an object with constant acceleration. Applying second kinematic equation for both the particles A and B along x and y axes, the angle between vector$\overline{\mathbf{a}}$ and${}^{\mathbf{+}\mathbf{y}}$axis can be found.

Formula:

The displacement in kinematic equation is given by,

$x=v_{0}t+\frac{1}{2}a{t}^2$ (i)

Where ${}^{v_0}$is the initial velocity

Equation of motion of particle A and particle B along y axis can be written using the second kinematic equation as,

$$\begin{gathered} \mathrm{y}={\mathrm{v}}_{\mathrm{y}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{y}}{\mathrm{t}}^2\mathrm{s} \\ 30\mathrm{m}=\frac{1}{2}\left[\left(0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\mathrm{³¦´Ç²õθ}\right]{\mathrm{t}}^2 \end{gathered}$$ (ii)

Equation of motion of particle A and particle B along x axis can be written using the second kinematic equation as,

$$\begin{gathered} \mathrm{x}={\mathrm{v}}_{\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2 \\ \left(3.0\mathrm{m}/\mathrm{s}\right)\mathrm{t}=\frac{1}{2}\left[\left(0.40\frac{\mathrm{m}}{{\mathrm{s}}^2}\right)\mathrm{²õ¾\pm ²Ôθ}\right]{\mathrm{t}}^2 \end{gathered}$$

Solving above equation for t,

$t=\frac{\left(3.0{\displaystyle \frac{m}{s}}\right)2}{\left(0.40{\displaystyle \frac{m}{s^2}}\right)\sin \mathrm{θ}}$ (iii)

Substituting equation (iii) in equation (ii),

$$\begin{gathered} 30m-\frac{1}{2}\left[\left(0.40\frac{m}{s^2}\right)\cos \mathrm{θ}\right]\left[\frac{2\left(3.0{\displaystyle \frac{m}{s}}\right)}{\left(0.40{\displaystyle \frac{m}{s^2}}\right)\sin \mathrm{θ}}\right] \\ As{\sin }^2\mathrm{θ}=1-{\cos }^2\mathrm{θ} \\ 1-{\cos }^2\mathrm{θ}-\frac{9.0}{\left(0.20\right)\left(30\right)}\cos \mathrm{θ} \\ {\cos }^2\mathrm{θ}-1.0\cos \mathrm{θ}-1-0 \\ \cos \mathrm{θ}-\frac{1}{2} \\ \mathrm{θ}={\cos }^{-1}\left(\frac{1}{2}\right) \\ \mathrm{θ}=60° \end{gathered}$$