Starting from the origin at time $t=0$, with initial velocity

Question:

Starting from the origin at time $t=0$, with initial velocity

$5 \hat{j} \mathrm{~ms}^{-1}$, a particle moves in the $x-y$ plane with a constant

acceleration of $(10 \hat{i}+4 \hat{j}) \mathrm{ms}^{-2}$. At time $t$, its coordiantes

are $\left(20 \mathrm{~m}, y_{0} \mathrm{~m}\right)$. The values of $t$ and $y_{0}$ are, respectively:

1. (1) $2 \mathrm{~s}$ and $18 \mathrm{~m}$

2. (2) $4 \mathrm{~s}$ and $52 \mathrm{~m}$

3. (3) $2 \mathrm{~s}$ and $24 \mathrm{~m}$

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

Correct Option: 1

Solution:

(1) Given: $\vec{u}=5 \hat{j} \mathrm{~m} / \mathrm{s}$

Acceleration, $\vec{a}=10 \hat{i}+4 \hat{j}$ and

final coordinate $\left(20, y_{0}\right)$ in time $t$.

$S_{x}=u_{x} t+\frac{1}{2} a_{x} t^{2}$                $\left[\because u_{x}=0\right]$

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

$S_{y}=u_{y} \times t+\frac{1}{2} a_{y} t^{2}$

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