A particle is moving in $X - Y p l a n e$ , which is initially at $x = y = 2 m$ . The particle's initial velocity is $2 {ms}^{- 1}$ along positive $X - a x i s$ and acceleration is $1 {ms}^{- 2}$ along positive $Y - a x i s$ . Determine the position of a particle at $t = 5 s$ .

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Solution

Step 1: Given data
The initial position of a particle at $X - a x i s$ , $x_{0} = 2 m$
The initial velocity of a particle along $X - a x i s$ , $u_{x} = 2 m s^{- 1}$
Acceleration of a particle along $X - a x i s$ , $a_{x} = 0 m s^{- 2}$
The initial position of a particle at $Y - a x i s$ , $y_{0} = 2 m$
The initial velocity of a particle along $Y - a x i s$ , $u_{y} = 0 m s^{- 1}$
Acceleration of a particle along $Y - a x i s$ , $a_{y} = 1 m s^{- 2}$
Time, $t = 5 s$
Step 2: Calculate the distance of a particle along $X - a x i s$ at $t = 5 s$
Using the second equation of the motion, $s = u t + \frac{1}{2} a t$ . Here, $u$ is the initial velocity of a particle, $s$ is the displacement of the particle, and $a$ is the acceleration of a particle at a time $t$ .
On putting the values along $X - a x i s$ in $s = u t + \frac{1}{2} a t$ :
$x = x_{0} + (u_{x} t + \frac{1}{2} a_{x} t)$
$= 2 + (2 \times 5 + \frac{1}{2} \times 0 \times 5^{2}) = 2 + (10 + 0) = 2 + 10 = 12 m$
Step 3: Calculate the distance of a particle along $Y - a x i s$ at $t = 5 s$
On putting the values along $Y - a x i s$ in $s = u t + \frac{1}{2} a t$ :
$y = y_{0} + (u_{y} t + \frac{1}{2} a_{y} t)$
$= 2 + (0 \times 5 + \frac{1}{2} \times 1 \times 5^{2}) = 2 + (0 + \frac{1}{2} \times 1 \times 25) = 2 + (0 + \frac{25}{2}) = 2 + \frac{25}{2} = \frac{2 \times 2 + 25}{2} = \frac{4 + 25}{2} = \frac{29}{2} = 14.5 m$
Hence, the position of a particle is $(12 m, 14.5 m)$ in $X - Y p l a n e$ at $t = 5 s$ .

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