Kinematics Ques 6
6 A particle moves from the point $(20 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathrm{m}$ at $t=0$ with an initial velocity $(5.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathrm{ms}^{-1}$. It is acted upon by a constant force which produces a constant acceleration $(4.0 \hat{\mathbf{i}}+4.0 \hat{\mathbf{j}}) \mathrm{ms}^{-2}$. What is the distance of the particle from the origin at time $2 \mathrm{~s}$ ?
(2019 Main, 11 Jan II)
(a) $5 \mathrm{~m}$
(b) $20 \sqrt{2} \mathrm{~m}$
(c) $10 \sqrt{2} \mathrm{~m}$
(d) $15 \mathrm{~m}$
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Answer:
Correct Answer: 6.( b )
Solution:
Formula:
- Given the initial position of the particle
$ r_0=(2 \hat{i}+4 \hat{j}) \mathrm{m}, $
Initial velocity of particle, $u=(5 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}$ Acceleration of particle, $a=(4 \hat{i}+5 \hat{j}) \mathrm{m} / \mathrm{s}^2$
According to the second equation of motion,
position of particle at time $t$ is, $\mathrm{r}=\mathrm{r}_0+\mathrm{u} t+\frac{1}{2} \mathrm{a} t^2$
At $t=2 \mathrm{~s}$, the position of the particle is,
$ \begin{aligned} & r =(2 \hat{i}+4 \hat{j})+(5 \hat{i}+4 \hat{j}) \times 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) \times 4 \\ \text { or } & r =(2+10+8) \hat{i}+(4+8+8) \hat{j} \\ \Rightarrow \quad & r =20 \hat{i}+20 \hat{j} \end{aligned} $
$\therefore$ Distance of the particle from the origin is,
$ |r|=20 \sqrt{2} \mathrm{~m} $
BETA
