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Physics · MOTION IN A PLANE
हिन्दीA particle starts from the origin at \( t = 0 \) with an initial velocity \( \mathbf{v}_0 = 5.0\hat{i}\text{ m/s} \) and moves in the \( x\text{-}y \) plane under a constant acceleration \( \mathbf{a} = (3.0\hat{i} + 2.0\hat{j})\text{ m/s}^2 \). When the \( x \)-coordinate of the particle is \( 84\text{ m} \), its \( y \)-coordinate and its speed are respectively:
Correct answer · Option 2
\( y = 36\text{ m} \), speed \( \approx 26\text{ m/s} \)
Explanation
\( x(t) = 5.0t + 1.5t^2 = 84 \implies 3t^2 + 10t - 168 = 0 \implies t = 6\text{ s} \). \( y(6) = \frac{1}{2}(2.0)(6)^2 = 36\text{ m} \). Velocity \( \mathbf{v}(6) = (5 + 3\times 6)\hat{i} + (2\times 6)\hat{j} = 23\hat{i} + 12\hat{j} \). Speed \( v = \sqrt{23^2 + 12^2} = \sqrt{673} \approx 26\text{ m/s} \).
