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The position of a particle is given by $\vec{r}=3.0 t \hat{i}-2.0 t^2 \hat{j}+4.0 \hat{k} m$, where $t$ is in seconds and the coefficients have the proper units for $\vec{r}$ to be in meters. (a) find the $v$ and $\vec{a}$ of the particle? (b) what is the magnitude and direction of velocity of the particle at $t=2.0 \mathrm{~s}$ ?
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Verified Answer
(a) Velocity, $\vec{v}=d r / d t$
$$
\begin{aligned}
&=\frac{d}{d t}\left(3.0 t \hat{i}-2.0 t^2 \hat{j}+4.0 t^2 \hat{k}\right) \\
&=3.0 \hat{i}-4.0 t \hat{j}
\end{aligned}
$$
Acceleration, $\vec{a}=d v / d t$
$$
=\frac{d}{d t}(3.0 \hat{i}-4.0 t \hat{j})=4.0 \hat{j}
$$
(b) At time $t=2 s$,
$$
\begin{aligned}
&\vec{v}=3.0 \hat{i}-4.0 \times 2 \hat{j}=3.0 \hat{i}-8.0 \hat{j} \\
&\left.v=\sqrt{[}(3)^2+(-8)^2\right] \mathrm{ms}^{-1}=\sqrt{73}=8.54 \mathrm{~ms}^{-1}
\end{aligned}
$$
If $\theta$ is the angle which $v$ makes with $x$-axis, then $\tan \theta=v_y / v_x=-8 / 3=-2.667=\tan 69.5^{\circ}$. $\therefore \theta=69.5^{\circ}$ below the $x$-axis.
$$
\begin{aligned}
&=\frac{d}{d t}\left(3.0 t \hat{i}-2.0 t^2 \hat{j}+4.0 t^2 \hat{k}\right) \\
&=3.0 \hat{i}-4.0 t \hat{j}
\end{aligned}
$$
Acceleration, $\vec{a}=d v / d t$
$$
=\frac{d}{d t}(3.0 \hat{i}-4.0 t \hat{j})=4.0 \hat{j}
$$
(b) At time $t=2 s$,
$$
\begin{aligned}
&\vec{v}=3.0 \hat{i}-4.0 \times 2 \hat{j}=3.0 \hat{i}-8.0 \hat{j} \\
&\left.v=\sqrt{[}(3)^2+(-8)^2\right] \mathrm{ms}^{-1}=\sqrt{73}=8.54 \mathrm{~ms}^{-1}
\end{aligned}
$$
If $\theta$ is the angle which $v$ makes with $x$-axis, then $\tan \theta=v_y / v_x=-8 / 3=-2.667=\tan 69.5^{\circ}$. $\therefore \theta=69.5^{\circ}$ below the $x$-axis.
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