JEE Mains · Physics · STD 11 - Motion in straight line
- JEE Mains 2024
- Hard
- Numerical
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___________.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(673\)
Step-by-step Solution
Detailed explanation
\( \mathrm{u}_{\mathrm{x}}=5 \mathrm{~m} / \mathrm{s} \quad \mathrm{a}_{\mathrm{x}}=3 \mathrm{~m} / \mathrm{s}^2 \quad \mathrm{x}=84 \mathrm{~m} \)
\( \mathrm{v}_{\mathrm{x}}^2-\mathrm{u}_{\mathrm{x}}^2=2 \mathrm{ax} \)
\( \mathrm{v}_{\mathrm{x}}^2-25=2(3)(84) \)
\( \mathrm{V}_{\mathrm{x}}=23 \mathrm{~m} / \mathrm{s} \)
\( \mathrm{v}_{\mathrm{x}}-\mathrm{u}_{\mathrm{x}}=\mathrm{a}_{\mathrm{x}} \mathrm{t} \)
\( \mathrm{t}=\frac{23-5}{3}=6 \mathrm{~s} \)
\( \mathrm{v}_{\mathrm{y}}=0+\mathrm{a}_{\mathrm{y}} \mathrm{t}=0+2 \times(6)=12 \mathrm{~m} / \mathrm{s} 4\)
\( \mathrm{v}^2=\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2=23^2+12^2=673 \)
\( \mathrm{v}=\sqrt{673} \mathrm{~m} / \mathrm{s}\)
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