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Q.

A particle starts from the origin at t = 0s with a velocity of $10.0 \hat{j} m / s$ and moves in the xy-plane with a constant acceleration of $(8.0 \hat{i} + 2.0 \hat{j}) {ms}^{- 2}$ . What time is the x-coordinate of the particle 16m?

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a

t = 2s

b

t = 3s

c

t = 4s

d

t = 1s

answer is A.

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Detailed Solution

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$\begin{array}{l} \bar{S} = \bar{ut} + \frac{1}{2} {\bar{at}}^{2} \\ \Rightarrow S_{x} \hat{i} + S_{y} \hat{j} = (10 \hat{j}) t + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) t^{2} \\ \Rightarrow S_{x} = 4 t^{2} = 16 \Rightarrow t^{2} = 4 \Rightarrow t = 2 s \end{array}$

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