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The position of a particle is given by $\vec{r}(t)=4t\hat{i}+2t^2\hat{j}+5\hat{k}$ where t is in seconds and r in metre. Find the magnitude and direction of velocity v(t), at t = 1s, with respect to x-axis
A
$4\sqrt2\ ms^{-1}$, $45^\circ$
B
$4\sqrt2\ ms^{-1}$, $60^\circ$
C
$3\sqrt2\ ms^{-1}$, $30^\circ$
D
$3\sqrt2\ ms^{-1}$, $45^\circ$
Detailed Solution
$\vec V=\dfrac{d\vec r}{dt}=4\hat i+4t\hat j+0\hat k$. At $t=1$ s, $\vec V=4\hat i+4\hat j$. $|\vec V|=\sqrt{4^2+4^2}=4\sqrt2$. $\tan\alpha=\dfrac{4}{4}=1\Rightarrow\alpha=45^\circ$.