Mathematics in Physics

JEE Main level, single correct, medium

34 of 121

Question 34 of 121

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} ?

Show answer and solution

Answer: (B) 202 m20 \sqrt{2} \mathrm{~m}

As S→=ut→+12a→t2\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{ut}}+\frac{1}{2} \overrightarrow{\mathrm{a}} \mathrm{t}^{2}

S→=(5i^+4j^)2+12(4i^+4j^)4\overrightarrow{\mathrm{S}}=(5 \hat{i}+4 \hat{j}) 2+\frac{1}{2}(4 \hat{i}+4 \hat{j}) 4

=10i^+8j^+8i^+8j^=10 \hat{i}+8 \hat{j}+8 \hat{i}+8 \hat{j}

r→f−r→i=18i^+16j^\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}=18 \hat{\mathrm{i}}+16 \hat{\mathrm{j}}

[ass→=\left[\mathrm{as} \overrightarrow{\mathrm{s}}=\right. change in position =r→f−r→i]\left.=\overrightarrow{\mathrm{r}}_{\mathrm{f}}-\overrightarrow{\mathrm{r}}_{\mathrm{i}}\right]

r→r=20i^+20j^\overrightarrow{\mathrm{r}}_{\mathrm{r}}=20 \hat{\mathrm{i}}+20 \hat{\mathrm{j}}

∣r→r∣=202\left|\overrightarrow{\mathrm{r}}_{\mathrm{r}}\right|=20 \sqrt{2}

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