Solve this following
Question:

Starting from the origin at time $t=0$, with initial velocity $5 \hat{\mathrm{j}} \mathrm{ms}^{-1}$, a particle moves in the $\mathrm{x}-\mathrm{y}$ plane with a constant acceleration of $(10 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}) \mathrm{ms}^{-2}$. At time $\mathrm{t}$, its coordinates are $\left(20 \mathrm{~m}, \mathrm{y}_{0} \mathrm{~m}\right)$. The values of $\mathrm{t}$ and $\mathrm{y}_{0}$, are respectively :

1. $4 \mathrm{~s}$ and $52 \mathrm{~m}$

2. $2 \mathrm{~s}$ and $24 \mathrm{~m}$

3. $2 \mathrm{~s}$ and $18 \mathrm{~m}$

4. $5 \mathrm{~s}$ and $25 \mathrm{~m}$

Correct Option: , 3

Solution:

Given $\quad \overrightarrow{\mathrm{u}}=5 \hat{\mathrm{j}} \mathrm{m} / \mathrm{s}, \overrightarrow{\mathrm{a}}=10 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}, \quad$ final

coordinate $\left(20, \mathrm{y}_{0}\right)$ in time $\mathrm{t}$

$\mathrm{S}_{\mathrm{x}}=4_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}$

$20-0=0+\frac{1}{2} \times 10 \times \mathrm{t}^{2}$

$\mathrm{t}=2 \mathrm{sec}$

$\mathrm{S}_{\mathrm{y}}=\mathrm{u}_{\mathrm{y}} \times \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}$

$y_{0}=5 \times 2+\frac{1}{2} 4 \times 2^{2}=18 m$

$2 \mathrm{sec}$ and $18 \mathrm{~m}$