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JEE Main 201911 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
11 January
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2019 · 11 January · Evening Shift · Physics · Mathematics in Physics

A particle moves from the point (2.0 i +4.0 j ) m, at t=0 , with an...

A particle moves from the point (2.0i^+4.0j^)m,(2.0 \hat{i}+4.0 \hat{j}) \mathrm{m}, at t=0\mathrm{t}=0, with an initial velocity (5.0i^+4.0j^)ms−1.(5.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-1} . It is acted upon by a constant force which produces a constant acceleration (4.0i^+4.0j^)ms−2.(4.0 \hat{i}+4.0 \hat{j}) \mathrm{ms}^{-2} . What is the distance of the particle from the origin at time 2 s?2 \mathrm{~s} ?

Options

  1. A15 m15 \mathrm{~m}
  2. B202 m20 \sqrt{2} \mathrm{~m}
  3. C5 m5 \mathrm{~m}
  4. D102 m10 \sqrt{2} \mathrm{~m}

Correct answer

B. 202 m20 \sqrt{2} \mathrm{~m}

Step-by-step solution

As S→=ut→+12a→t2S→=(5i^+4j^)2+12(4i^+4j^)4=10i^+8j^+8i^+8j^r→f−r→i=18i^+16j^[ass→=change\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{ut}}+\frac{1}{2} \overrightarrow{\mathrm{a}} \mathrm{t}^{2} \overrightarrow{\mathrm{S}}=(5 \hat{i}+4 \hat{j}) 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) 4 =10 \hat{i}+8 \hat{j}+8 \hat{i}+8 \hat{j} \overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}=18 \hat{\mathrm{i}}+16 \hat{\mathrm{j}} \left[\mathrm{as} \overrightarrow{\mathrm{s}}=\right.change in position \left.=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}\right]\overrightarro\overrightarrow{\mathrm{r}}_{\mathrm{r}}=20 \hat{\mathrm{i}}+20 \hat{\mathrm{j}} \left|\overrightarrow{\mathrm{r}}_{\mathrm{r}}\right|=20 \sqrt{2}
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