Question

A particle starts from the origin at $t=0$with a velocity of $5m{s}^{-1}$ along +ve X-axis and moves in X-Y plane under action of constant acceleration $\overrightarrow{a}=(3\hat{i}+2\hat{j})m{s}^{-2}$. Determine the

(a) Y-coordinate of the particle at the instant when its X-coordinate is $84m$.

(b) speed of the particle at this moment.

Answer 1

Step 1: Given data

The $x$ component of the initial velocity, $u_x=5m{s}^{-1}$.

The $y$ component of the initial velocity, $u_y=0m{s}^{-1}$.

The $x$ component of the acceleration, $a_x=3m{s}^{-1}$.

The $y$ component of the acceleration, $a_y=2m{s}^{-1}$.

Step 2: Find the time taken by the particle to reach the $\mathit{X}\mathbf{-}$coordinate equal to $\mathbf{84}\mathbf{}\mathit{m}$.

Using the second equation of motion, the $x$ component of the displacement is $\mathit{X}\mathbf{}\mathbf{=}\mathbf{}{\mathit{u}}_{\mathbf{x}}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{x}}{\mathit{t}}^{\mathbf{2}}$.

Here, $X$ is the distance covered along the $x-axis$ and $t$ is the time taken for covering that distance.

$\Rightarrow X=5t+\frac{1}{2}3t^2$

$\Rightarrow 84=\frac{10t+3t^2}{2}$

$\Rightarrow 3t^2+10t-168=0$

Using the formula $\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{\left(}\mathbf{b}\mathbf{\right)}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}\mathbf{}$ to find the roots of the above quadratic equation,

$$\begin{gathered} \Rightarrow t=\frac{-10+\sqrt{{\left(10\right)}^2-4\left(3\right)\left(168\right)}}{2\left(3\right)}andt=\frac{-10-\sqrt{{\left(10\right)}^2-4\left(3\right)\left(168\right)}}{2\left(3\right)} \\ \Rightarrow t=\frac{-10+\sqrt{2116}}{6}andt=\frac{-10-\sqrt{2116}}{6} \end{gathered}$$

$$\begin{gathered} \Rightarrow t=\frac{-10+46}{6}andt=\frac{-10-46}{6} \\ \Rightarrow t=\frac{36}{6}andt=\frac{-56}{6} \\ \Rightarrow t=6sandt=-9.33s \end{gathered}$$

As time cannot be negative that is why $t=6s$.

Step 3: Find the $\mathit{Y}\mathbf{-}$coordinate of the particle at the instant its $\mathit{X}\mathbf{-}$coordinate is $\mathbf{84}\mathbf{}\mathit{m}$.

Now, using the second equation of motion, the $y$ component of the displacement is $\mathit{Y}\mathbf{}\mathbf{=}\mathbf{}{\mathit{u}}_{\mathbf{y}}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{a}}_{\mathbf{y}}{\mathit{t}}^{\mathbf{2}}$

Here, $Y$ is the distance covered along the $y-axis$ and $t$ is the time taken for covering that distance.

$\Rightarrow Y=0+\frac{1}{2}\left(2\right){\left(6\right)}^2[\mathit{t}\mathbf{=}\mathbf{6}\mathit{s}(\mathrm{calculated}\mathrm{in}\mathrm{step}2\left)\right]$

$\Rightarrow Y=36m$

Step 4: Find the speed of the particle at $\mathbf{6}\mathbf{}\mathit{s}$.

Using the first equation of motion, the $x$ component of the velocity is ${\mathit{v}}_{\mathbf{x}}\mathbf{=}\mathbf{}{\mathit{u}}_{\mathbf{x}}\mathbf{+}{\mathit{a}}_{\mathbf{x}}\mathit{t}$

$\Rightarrow v_x=5+3\times 6$

$\Rightarrow v_x=23m{s}^{-1}$

Now, the $y$ component of the velocity is ${\mathit{v}}_{\mathbf{y}}\mathbf{=}\mathbf{}{\mathit{u}}_{\mathbf{y}}\mathbf{+}{\mathit{a}}_{\mathbf{y}}\mathit{t}$

$\Rightarrow v_y=0+2\times 6$

$\Rightarrow v_y=12m{s}^{-1}$

The velocity vector or the speed of the particle at this moment is:

$\stackrel{\rightharpoonup }{v}=\sqrt{{\left(v_x\right)}^2+{\left(v_y\right)}^2}$

$$\begin{gathered} \stackrel{\rightharpoonup }{v}=\sqrt{{\left(23\right)}^2+{\left(12\right)}^2} \\ \stackrel{\rightharpoonup }{v}=\sqrt{529+144} \end{gathered}$$

$\stackrel{\rightharpoonup }{v}=25.94m{s}^{-1}$

Thus:

(a) The $Y-$coordinate of the particle is $\mathbf{36}\mathbf{}\mathit{m}$ at the instant its $X-$coordinate is $84m$.

(b) The speed of the particle when the $X-$coordinate is $\mathbf{25}\mathbf{.}\mathbf{94}\mathbf{}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.