question_answer

Two particles start simultaneously from the same point and move along two straight lines, one with uniform velocity \[\vec{u}\] and the other from rest with uniform acceleration\[\vec{f}\]. Let \[\alpha \] be the angle between their directions of motion. The relative velocity of the second particle w.r.t. the first is least after a time        

A) \[\frac{u\,\,\cos \,\,\alpha }{f}\]

B) \[\frac{u\,\,\sin \,\,\alpha }{f}\]

C) \[\frac{f\,\,\cos \,\,\alpha }{u}\]

D) \[u\,\,\sin \,\,\alpha \]

Correct Answer: A

Solution :

[a] We can consider the two velocities as \[{{\vec{v}}_{1}}=u\hat{i}\] and \[{{\vec{v}}_{2}}=(ft\,\,cos\,\,\alpha )\hat{i}+(ft\,\,sin\,\,\alpha )\hat{j}\] \[\therefore \]Relative velocity of Second with respect to first \[\vec{v}={{\vec{v}}_{2}}-{{\vec{v}}_{1}}=(ft\,\,\cos \,\,\alpha -u)i+ft\,\,\sin \,\,\alpha \,\,\hat{j}\] \[\Rightarrow {{\left| {\vec{v}} \right|}^{2}}={{(ft\,\,\cos \,\,\alpha -u)}^{2}}+{{(ft\,\,\sin \,\,\alpha )}^{2}}\]             \[{{f}^{2}}{{t}^{2}}+{{u}^{2}}-2uft\,\,\cos \alpha \] For \[\left| {\vec{v}} \right|\] to be min we should have \[\frac{d{{\left| v \right|}^{2}}}{dt}=0\Rightarrow 2{{f}^{2}}t-2uf\,\,\cos \,\,\alpha =0\] \[\Rightarrow t=\frac{u\,\,\cos \,\,\alpha }{f}\] Also      \[\frac{{{d}^{2}}{{\left| v \right|}^{2}}}{d{{t}^{2}}}=2{{f}^{2}}=+ve\] \[\therefore {{\left| v \right|}^{2}}\]and hence \[\left| v \right|\] is least at the time \[\frac{u\,\,\cos \,\,\alpha }{f}\]