question_answer

A body of mass 2 kg has an initial velocity of \[\text{5 m}{{\text{s}}^{-1}}\] along OE and it is subjected to a force of 4 N in a direction perpendicular to \[\text{OE}.\]The distance of the body from 0 after 4 s will be

A)  20m                                     

B)  48m

C)  26m                                     

D)  12m

Correct Answer: C

Solution :

                 From Newtons law F = ma  where m is mass and a the acceleration.                                                   Given,   F = 4 N, m = 2 kg \[\therefore \]  \[a=\frac{F}{m}=\frac{4}{2}=2\,m{{s}^{-2}}\] Also, distance covered by body is                 \[d=4\times 5=20\,m\] From equation of motion, displacement (y) is given by                 \[y=ut+\frac{1}{2}a{{t}^{2}}\] Given,   u = 0;     \[a=2\,m{{s}^{-2}},\,\,\,t=4\,s\]                 \[\therefore \]  \[y=0+\frac{1}{2}\times 2\times {{(4)}^{2}}=16\,\,m\] In vector form, displacement is given by                 \[s=x\,\hat{i}+y\,\hat{j}\]                 \[s=20\,\hat{i}+16\,\hat{j}\] Hence, distance from 0 is                 \[\left| s \right|=\sqrt{{{20}^{2}}+{{16}^{2}}}\]                 \[\left| s \right|=\sqrt{400+256}\]                 \[\left| s \right|=25.6\,m\] \[\Rightarrow \]               \[s\approx 26\,\,m\]