A) \[\text{18\hat{i} + 3\hat{j}}\]
B) \[\text{18\hat{i} + 6\hat{j}}\]
C) \[\text{3\hat{i} + 18\hat{j}}\]
D) \[\text{18\hat{i} + 4\hat{j}}\]
Correct Answer: B
Solution :
Key Idea: Force applied on the object is rate of change of momentum. According to Newton's 2nd law, force applied on an object is equal to rate of change of momentum. That is, \[\vec{F}=\frac{d\,\vec{p}}{dt}\] or \[\vec{F}=m\frac{d\,\vec{v}}{dt}.....(i)\] Given, m = 3 kg, \[t=3\,s,\,\vec{F}=(6{{t}^{2}}\,\hat{i}+4t\hat{j})\,n\] Substituting theses values in Eq. (i) we get \[(6{{t}^{2}}\,\hat{i}+4t\,\hat{j})=3\,\frac{d\,\vec{v}}{dt}\] \[ord\,\vec{v}=\frac{1}{3}(6{{t}^{2}}\hat{i}+4t\hat{j})\,dt\] Now, taking integration of both sides, we get \[\int{d\,\vec{v}=\int\limits_{0}^{t}{\frac{1}{3}}\,\,(6{{t}^{2}}\hat{i}+4t\hat{j})\,d}t\] \[\vec{v}=\frac{1}{3}\,\int\limits_{0}^{t}{(6{{t}^{2}}\hat{i}+4t\hat{j})\,dt}\] \[butt=3\,s\,(given)\] \[\therefore \vec{v}=\frac{1}{3}\,\int\limits_{0}^{3}{\,\,(6{{t}^{2}}\hat{i}+4t\hat{j})\,dt}\] \[or\vec{v}=\frac{1}{3}\,\left[ \frac{6\,{{t}^{3}}}{3}\hat{i}+\frac{4\,{{t}^{2}}}{2}\,\hat{j} \right]_{0}^{3}\] \[orv=\frac{1}{3}\,[2{{(3)}^{3}}\hat{i}+2{{(3)}^{2}}\hat{j}]\] or \[\vec{v}=\frac{1}{3}[54\,\hat{i}+18\hat{j}]\] or \[\vec{v}=18\hat{i}+6\hat{j}\]
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