JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2020
- Medium
- MCQ
A particle starts from the origin at \(\mathrm{t}=0\) with an initial velocity of \(3.0 \hat{\mathrm{i}} \;\mathrm{m} / \mathrm{s}\) and moves in the \(x-y\) plane with a constant acceleration \((6.0 \hat{\mathrm{i}}+4.0 \hat{\mathrm{j}}) \;\mathrm{m} / \mathrm{s}^{2} .\) The \(\mathrm{x}\) -coordinate of the particle at the instant when its \(y-\)coordinate is \(32\;\mathrm{m}\) is \(D\) meters. The value of \(D\) is
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Answer & Solution
Correct Answer
(C) \(60\)
Step-by-step Solution
Detailed explanation
\(\mathrm{x}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}\)
\(\mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}\)
\(32=0 \times t+\frac{1}{2}(4)(t)^{2}\)
\(t^{2}=16\)
\(t=4 \mathrm{sec}\)
\(x=3 \times 4+\frac{1}{2} \times 6 \times 4^{2}\)
\(=12+48=60 \mathrm{m}\)
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