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A particle of mass $m=1 \mathrm{~kg}$ moves in the $x y$-plane. The force on it at time $t$ is $F(t)=[2 \sin (\alpha t) \hat{\mathbf{i}}+3 \cos (\alpha t) \hat{\mathbf{j}}] \mathrm{N}$, where $\alpha=1 \mathrm{~s}^{-1}$.
At time $t=0$, the particle is at rest at the origin. Calculate the magnitude of its position vector $\mathbf{r}$ (in $\mathrm{m}$ ) and velcoity vector $\mathbf{v}$ (in $\mathrm{m} / \mathrm{s}$ ) at time $t=\frac{\pi}{2} \mathrm{~s}$.
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At time $t=0$, the particle is at rest at the origin. Calculate the magnitude of its position vector $\mathbf{r}$ (in $\mathrm{m}$ ) and velcoity vector $\mathbf{v}$ (in $\mathrm{m} / \mathrm{s}$ ) at time $t=\frac{\pi}{2} \mathrm{~s}$.
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The correct answer is:
None of these
Given, $m=1 \mathrm{~kg}$
Force, $F(t)=[2 \sin \alpha t \hat{\mathbf{i}}+3 \cos \alpha t \hat{\mathbf{j}}] \mathrm{N}$
Here, $\quad \alpha=1 \mathrm{~s}^{-1} \Rightarrow F(t)=(2 \sin t \hat{\mathbf{i}}+3 \cos t \hat{\mathbf{j}}) \mathrm{N}$
So, component of force,
$F_x=2 \sin t \text { and } F_y=3 \cos t$
As, $\quad F_x=\frac{m d v_x}{d t}=2 \sin t$
$\Rightarrow \quad d v_x=2 \sin t d t \quad(\because m=1 \mathrm{~kg})$
Integrating both sides, we get
$v_x=\int_0^{\pi / 2} 2 \sin t d t=-2[\cos t]_0^{\pi / 2}$
$=-2\left(\cos \frac{\pi}{2}-\cos 0\right)=-2(0-1)=2 \mathrm{~m} / \mathrm{s}$
Similarly, $v_y=\int_0^{\pi / 2} 3 \cos t d t=3[\sin t]_0^{\pi / 2}$
$=3\left(\sin \frac{\pi}{2}-\sin 0\right)=3(1-0)=3 \mathrm{~m} / \mathrm{s}$
$\therefore$ Magnitude of velocity vector,
$|v|=\sqrt{v_x^2+v_y^2}=\sqrt{(2)^2+(3)^2}=\sqrt{13} \mathrm{~m} / \mathrm{s}$
As, velocity vector,
$\mathbf{v}=v_x \hat{\mathbf{i}}+v_y \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$
So,
$v_x=\frac{d r_x}{d t}=2 \mathrm{~m} / \mathrm{s} \text { and } v_y=\frac{d r_y}{d t}=3 \mathrm{~m} / \mathrm{s}$
$\Rightarrow \quad d r_x=2 d t$
Integrating both sides, we get
$r_x=\int_0^{\pi / 2} 2 d t=2[t]_0^{\pi / 2}=\pi \mathrm{m}$
Similarly, $r_y=\int_0^{\pi / 2} 3 d t=3[t]_0^{\pi / 2}=\frac{3 \pi}{2} \mathrm{~m}$
$\therefore$ Magnitude of position vector,
$|r|=\sqrt{r_x^2+r_y^2}=\sqrt{\pi^2+\left(\frac{3 \pi}{2}\right)^2}=\sqrt{13} \frac{\pi}{2} \mathrm{~m}$
Force, $F(t)=[2 \sin \alpha t \hat{\mathbf{i}}+3 \cos \alpha t \hat{\mathbf{j}}] \mathrm{N}$
Here, $\quad \alpha=1 \mathrm{~s}^{-1} \Rightarrow F(t)=(2 \sin t \hat{\mathbf{i}}+3 \cos t \hat{\mathbf{j}}) \mathrm{N}$
So, component of force,
$F_x=2 \sin t \text { and } F_y=3 \cos t$
As, $\quad F_x=\frac{m d v_x}{d t}=2 \sin t$
$\Rightarrow \quad d v_x=2 \sin t d t \quad(\because m=1 \mathrm{~kg})$
Integrating both sides, we get
$v_x=\int_0^{\pi / 2} 2 \sin t d t=-2[\cos t]_0^{\pi / 2}$
$=-2\left(\cos \frac{\pi}{2}-\cos 0\right)=-2(0-1)=2 \mathrm{~m} / \mathrm{s}$
Similarly, $v_y=\int_0^{\pi / 2} 3 \cos t d t=3[\sin t]_0^{\pi / 2}$
$=3\left(\sin \frac{\pi}{2}-\sin 0\right)=3(1-0)=3 \mathrm{~m} / \mathrm{s}$
$\therefore$ Magnitude of velocity vector,
$|v|=\sqrt{v_x^2+v_y^2}=\sqrt{(2)^2+(3)^2}=\sqrt{13} \mathrm{~m} / \mathrm{s}$
As, velocity vector,
$\mathbf{v}=v_x \hat{\mathbf{i}}+v_y \hat{\mathbf{j}}=2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$
So,
$v_x=\frac{d r_x}{d t}=2 \mathrm{~m} / \mathrm{s} \text { and } v_y=\frac{d r_y}{d t}=3 \mathrm{~m} / \mathrm{s}$
$\Rightarrow \quad d r_x=2 d t$
Integrating both sides, we get
$r_x=\int_0^{\pi / 2} 2 d t=2[t]_0^{\pi / 2}=\pi \mathrm{m}$
Similarly, $r_y=\int_0^{\pi / 2} 3 d t=3[t]_0^{\pi / 2}=\frac{3 \pi}{2} \mathrm{~m}$
$\therefore$ Magnitude of position vector,
$|r|=\sqrt{r_x^2+r_y^2}=\sqrt{\pi^2+\left(\frac{3 \pi}{2}\right)^2}=\sqrt{13} \frac{\pi}{2} \mathrm{~m}$
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