In Fig. 4-32, particle${}^{\mathbf{A}}$moves along the linelocalid="1654157716964" ${}^{\mathit{y}\mathbf{=}\mathbf{30}\mathbf{}\mathit{m}}$with a constant velocitylocalid="1654157724548" $\stackrel{\mathbf{\rightharpoonup }}{\mathit{v}}$of magnitude localid="1654157733335" ${}^{\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}}$and parallel to thelocalid="1654157740474" ${}^{\mathbf{x}}$axis. At the instant particlelocalid="1654157754808" ${}^{\mathbf{A}}$passes thelocalid="1654157747127" ${}^{\mathbf{y}}$axis, particlelocalid="1654157760143" $\mathit{B}$leaves the origin with a zero initial speed and a constant accelerationlocalid="1654157772700" ${}^{\mathit{a}}$of magnitudelocalid="1654157766903" $\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What anglelocalid="1654157780936" ${}^{\mathbf{\theta }}$between $\overline{\mathbf{a}}$and the positive direction of thelocalid="1654157787877" ${}^{\mathbf{y}}$axis would result in a collision?

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 ${}^{\stackrel{\rightharpoonup }{\mathrm{a}}=0.40m/s^2}$

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,

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

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

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

Solving above equation for t,

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

Substituting equation (iii) in equation (ii),

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