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JEE Mains · Physics · STD 11 - 3.2 motion in plane

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

  1. A \(50\)
  2. B \(32\)
  3. C \(60\)
  4. D \(40\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(60\)

Step-by-step Solution

Detailed explanation

\(\mathrm{x}=\mathrm{u}_{\mathrm{x}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{x}} \mathrm{t}^{2}\) \(\mathrm{y}=\mathrm{u}_{\mathrm{y}} \mathrm{t}+\frac{1}{2} \mathrm{a}_{\mathrm{y}} \mathrm{t}^{2}\) \(32=0 \times t+\frac{1}{2}(4)(t)^{2}\) \(t^{2}=16\) \(t=4 \mathrm{sec}\)…
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