Question

A particle starts from origin with an initial velocity of $5{\mathrm{ms}}^{-1}$ along positive $Y-axis$, under a constant acceleration whose $X$ and $Y$ components are $2{\mathrm{ms}}^{-2}$ and $1{\mathrm{ms}}^{-2}$, respectively. Determine the velocity (magnitude and direction both) of the particle at $t=4\mathrm{s}$.

Answer 1

Step 1: Given data

The initial velocity of a particle along $Y-axis$ , $u_y=5m{s}^{-1}$.

The initial velocity of a particle along $X-axis$ , $u_x=0m{s}^{-1}$.

Acceleration of a particle along $X-axis$ , $a_x=2m{s}^{-2}$.

Acceleration of a particle along $Y-axis$ , $a_y=1m{s}^{-2}$.

Time, $t=4s$

Step 2: Calculate the velocity of the particle along $\mathit{Y}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$ at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathit{s}$

The formula used: $v=u+at$.

Here, $v$ is the velocity of the particle at time $t$, $u$ is the initial velocity of the particle and $a$ is the acceleration of a particle at time $t$.

$\therefore v_y=u_y+a_{y}t$

$$\begin{gathered} =5+1\times 4 \\ =5+4 \\ =9m{s}^{-1} \end{gathered}$$

Step 3: Calculate the velocity of the particle along $\mathit{X}\mathbf{-}\mathit{a}\mathit{x}\mathit{i}\mathit{s}$ at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathit{s}$

The formula used: $v=u+at$.

$\therefore v_x=u_x+a_{x}t$

$$\begin{gathered} =0+2\times 4 \\ =0+8 \\ =8m{s}^{-1} \end{gathered}$$

Step 4: Calculate the magnitude of the velocity at $\mathit{t}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{}\mathit{s}$

The value of $v_x=8m{s}^{-1}$ and $v_y=9m{s}^{-1}$.

$\therefore v=v_x\hat{i}+v_y\hat{j}$
$=\left(8\hat{i}+9\hat{j}\right)m{s}^{-1}$

Hence, the velocity of the particle at $t=4s$ is ${\left(8\hat{i}+9\hat{j}\right)}^{\mathbf{}}\mathit{m}{\mathit{s}}^{\mathbf{-}\mathbf{1}}$.