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JEE Mains · Physics · STD 11 - 2. motion in straight line

A particle starts from origin at \(t=0\) with a velocity \(5 \hat{i} \mathrm{~m} / \mathrm{s}\) and moves in \(x-y\) plane under action of a force which produces a constant acceleration of \((3 \hat{i}+2 \hat{j}) \mathrm{m} / \mathrm{s}^2\). If the \(x\)-coordinate of the particle at that instant is \(84 \mathrm{~m}\), then the speed of the particle at this time is \(\sqrt{\alpha} \mathrm{m} / \mathrm{s}\). The value of \(\alpha\) is___________.

  1. A \(673\)
  2. B \(685\)
  3. C \(756\)
  4. D \(741\)
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Answer & Solution

Correct Answer

(A) \(673\)

Step-by-step Solution

Detailed explanation

\( \mathrm{u}_{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}_{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^2 \quad \mathrm{x}=84 \mathrm{~m} \) \( \mathrm{v}_{\mathrm{x}}^2-\mathrm{u}_{\mathrm{x}}^2=2 \mathrm{ax} \) \( \mathrm{v}_{\mathrm{x}}^2-25=2(3)(84) \)…
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