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The position of a particle is given by $\vec{r} = (3 t \hat{i} + 2 t^{2} \hat{j} + 5 \hat{k})$ . Where ‘t’ is in seconds and the coefficients have proper units for ‘r’ to be in metre. The velocity of particle at t = 1s is $(\sin 37^{\circ} = 0.6)$

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$5 m s^{- 1}, 37^{0}$ with x-axis

$5 m s^{- 1}, 53^{0}$ with x-axis

$4 m s^{- 1}, 37^{0}$ with x-axis

$4 m s^{- 1}, 53^{0}$ with x-axis

answer is B.

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Detailed Solution

$V = \frac{\vec{d r}}{d t} = (3 \hat{i} + 4 t \hat{j})$

At t = 1s, $\vec{V} = (3 \hat{i} + 4 \hat{j})$

$\Rightarrow V = 5 m s^{- 1}$ and 53⁰ with + Ve x-axis

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