Question

A particle is moving in $X-Yplane$, which crosses origin at $t=0$ with a velocity of $10{\mathrm{ms}}^{-1}$ along positive $X-axis$. A constant acceleration, whose $X$ and $Y$ components are $-2{\mathrm{ms}}^{-2}$ and $1{\mathrm{ms}}^{-2}$ respectively is acting on particle. Determine
(a) the time when velocity vector becomes parallel to $Y-axis$.
(b) the position of the particle in above situation.
(c) the time when the particle crosses $Y-axis$.
(d) the speed of the particle when it crosses $Y-axis$.

Answer 1

Step 1: Given data

The initial velocity at $t=0s$ along positive $X-axis$, $u_x=10m{s}^{-1}$

The acceleration at $t=0s$ along $X-axis$, $a_x=-2m{s}^{-2}$

The initial velocity at $t=0s$ along positive $Y-axis$, $u_y=0m{s}^{-1}$

The acceleration at $t=0s$ along $Y-axis$, $a_y=1m{s}^{-2}$

Step 2: Find the velocity vector of a particle

The $x$ component of the velocity of a particle, $v_x=u_x+a_{x}t$

$$\begin{gathered} =10+(-2)t \\ =\left(10-2t\right)m{s}^{-1}..........\left(1\right) \end{gathered}$$

The $y$ component of the velocity of a particle, $v_y=u_y+a_{y}t$

$$\begin{gathered} =0+\left(1\right)t \\ =tm{s}^{-1} \end{gathered}$$

∴ The velocity vector, $\stackrel{\rightharpoonup }{v}=v_x\hat{i}+v_y\hat{j}$

$=\left[\left(10-2t\right)\hat{i}+t\hat{j}\right]m{s}^{-1}............\left(2\right)$

Step 3: Find the position vector of a particle

The $x$ component of the position of a particle, $r_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$

$$\begin{gathered} =10t+\frac{1}{2}(-2)t^2 \\ =\left(10t-t^2\right)m..............\left(3\right) \end{gathered}$$

The $y$ component of the position of a particle, $r_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$

$$\begin{gathered} =0\times t+\frac{1}{2}\times 1\times t^2 \\ =0+\frac{t^2}{2} \\ =\frac{t^2}{2}m \end{gathered}$$

∴ The position vector, $\stackrel{\rightharpoonup }{r}=r_x\hat{i}+r_y\hat{j}$

$=\left[\left(10t-t^2\right)\hat{i}+\frac{t^2}{2}\hat{j}\right]m...........\left(4\right)$

Step 4: Find the time at which the velocity vector of a particle becomes parallel to the $\mathit{Y}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$

(a) The velocity vector becomes parallel to the $Y-axis$ when the $x$ component of the velocity of a particle becomes equal to zero.

$$\begin{gathered} \Rightarrow v_x=0 \\ \Rightarrow 10-2t=0\left[Fromequation\left(1\right)\right] \\ \Rightarrow 2t=10 \\ \Rightarrow t=\frac{10}{2} \\ \Rightarrow t=5s \end{gathered}$$

Step 5: Find the position vector of a particle at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{5}\mathbf{}\mathit{s}$

(b) Substitute the value of $t$ as $5s$ in the equation $\left(4\right)$

∴ The position vector at $5s$ $=\left(10t-t^2\right)\hat{i}+\frac{t^2}{2}\hat{j}=\left(10\times 5+5^2\right)\hat{i}+\frac{5^2}{2}\hat{j}$

$$\begin{gathered} =\left(50-25\right)\hat{i}+\frac{25}{2}\hat{j} \\ =\left(25\hat{i}+12.5\hat{j}\right)m \end{gathered}$$

Step 6: Find the time when the particle crosses the $\mathit{Y}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$

(c) The $x$ component of the position of a particle will be equal to zero when the particle crosses the $Y-axis$

$$\begin{gathered} \Rightarrow r_x=0 \\ \Rightarrow 10t-t^2=0\left[Fromequation\left(3\right)\right] \\ \Rightarrow t\left(10-t\right)=0 \\ \Rightarrow 10-t=0 \\ \Rightarrow t=10s \end{gathered}$$

Step 7: Find the speed of a particle when it crosses the $\mathit{Y}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$

(d) As we have calculated the time at which a particle crosses the $Y-axis$ is $t=10s$ in Step 6.

∴ The velocity vector at $10s$ $=\left(10-2\times 10\right)\hat{i}+10\hat{j}\left[Fromequation\left(2\right)\right]$

$$\begin{gathered} =\left(10-20\right)\hat{i}+10\hat{j} \\ =\left(-10\hat{i}+10\hat{j}\right)m{s}^{-1} \end{gathered}$$

Final Answer:

(a) The time when the velocity vector becomes parallel to $Y-axis$ is $\mathbf{5}\mathbf{}\mathit{s}$.
(b) The position of the particle when the velocity vector becomes parallel to $Y-axis$ is $\left(25m,12.5m\right)$.
(c) The time when the particle crosses $Y-axis$ is $\mathbf{10}\mathbf{}\mathit{s}$.
(d) The speed of the particle when the particle crosses $Y-axis$ is $\mathbf{10}\sqrt{\mathbf{2}}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.