Q170 / 180 · COMEDK 2015 · Physics
A body of mass $1 \mathrm{~kg}$ starting from origin with initial velocity of $3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}} \mathrm{ms}^{-1}$. If a constant force of $2 \hat{\mathbf{i}}-3 \hat{\mathfrak{j}}$ acts on it, find the coordinate of point it reaches in $3 \mathrm{~s}$.
- Correct answerA$(18,-1.5)$
- B$(9,1.5)$
- C$(18,1.5)$
- D$(9,-1.5)$
Solution
Given, initial velocity, $u=3 \hat{\mathrm{i}}+4 \hat{\hat{\mathrm{s}}}$
Force, $F=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}$
Initial velocity along $X$-axis and $Y_{\text {-axis }}$ are $u_{x}=3 \mathrm{~ms}^{-1}$
and $u_{y}=4 \mathrm{~ms}^{-1}$, respectively.
Similarly, $F_{x}=2 \mathrm{~N}$ and $F_{y}=-3 \mathrm{~N}$
If acceleration along $X$ and $Y$-axes are $a_{x}$ and $a_{y}$, then
$$
a_{x}=\frac{F_{x}}{m}=\frac{2}{1}=2 \mathrm{~m} / \mathrm{s}^{2} \quad[\because F=m a]
$$
$$
\begin{aligned}
a_{y} &=\frac{F_{y}}{m}=\frac{-3}{1}=-3 \mathrm{~m} / \mathrm{s}^{2} \\
s_{x} &=u_{x} t+\frac{1}{2} u_{x} l^{2}=3 \times 3+\frac{1}{2} \times 2 \times 9=18 \mathrm{un} \\
s_{y} &=u_{y} t+\frac{1}{2} a_{y} t^{2}=4 \times 3+\frac{1}{2} \times(-3) \times 9 \\
&=12-13.5=-1.5 \mathrm{~m}
\end{aligned}
$$
So, required $x$ and $y$-coordinates of point $=(18,-15)$
Hence, the correct answer is option: 'A'.