A particle moves along the positive branch of the curve $y = \frac{x}{2}$ where $x = \frac{t^{2}}{2}$ where x and y are measured in meter and t in second. At t = 2 sec, the velocity of the particle is:

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$(4 \hat{i} - 2 \hat{j}) m / \sec$

$(2 \hat{i} - 4 \hat{j}) m / \sec$

$(2 \hat{i} + 4 \hat{j}) m / \sec$

$(2 \hat{i} + 2 \hat{j}) m / \sec$

answer is B.

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Detailed Solution

As $y = \frac{x}{2}$ where $x = \frac{t^{2}}{2}$ so, $v_{x} = \frac{2 t}{2} = t and v_{y} = \frac{2 {xv}_{x}}{2} = \frac{t^{2}}{2} (t) = \frac{t^{3}}{2}$
Therefore at t = 2s
$\vec{v} = (2 \hat{i} + 4 \hat{j}) m / \sec$

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