JEE Mains · Physics · STD 11 - Motion in plane
- JEE Mains 2019
- Medium
- MCQ
A particle moves from the point \(\left( {2.0\hat i + 4.0\hat j} \right)\,m\), at \(t = 0\) with an initial velocity \(\left( {5.0\hat i + 4.0\hat j} \right)\,m{s^{ - 1 }}\). It is acted upon by a constant force which produces a constant acceleration \(\left( {4.0\hat i + 4.0\hat j} \right)\,m{s^{ - 2}}\). What is the distance of the particle from the origin at time \(2\,s\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(20\sqrt 2 \,m\)
Step-by-step Solution
Detailed explanation
\(\begin{array}{l}
\overrightarrow S = \left( {5\hat i + 4\hat j} \right)2 + \frac{1}{2}\left( {4\hat j + 4\hat j} \right)4\\
\,\,\,\, = \,10\hat i + 8\hat j + 8\hat i + 8\hat j\\
\overrightarrow {{r_2}} - \overrightarrow {{r_1}} = 18\hat i + 16\hat j\\
\overrightarrow {{r_2}} = 20\hat i + 20\hat j\\
\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2
\end{array}\)
\overrightarrow S = \left( {5\hat i + 4\hat j} \right)2 + \frac{1}{2}\left( {4\hat j + 4\hat j} \right)4\\
\,\,\,\, = \,10\hat i + 8\hat j + 8\hat i + 8\hat j\\
\overrightarrow {{r_2}} - \overrightarrow {{r_1}} = 18\hat i + 16\hat j\\
\overrightarrow {{r_2}} = 20\hat i + 20\hat j\\
\left| {\overrightarrow {{r_f}} } \right| = 20\sqrt 2
\end{array}\)
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