Vector Algebra Question 28
Question: 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
Options:
A) $ \frac{u\cos \alpha }{f} $
B) $ \frac{u\sin \alpha }{f} $
C) $ \frac{f\cos \alpha }{u} $
D) $ u\sin \alpha $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] We can consider the two velocities as $ {{\vec{v}}_1}=u\hat{i} $ and $ {{\vec{v}}_2}=(ftcos\alpha )\hat{i}+(ftsin\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 {{| {\vec{v}} |}^{2}}={{(ft\cos \alpha -u)}^{2}}+{{(ft\sin \alpha )}^{2}} $ $ f^{2}t^{2}+u^{2}-2uft\cos \alpha $ For $ | {\vec{v}} | $ to be min we should have $ \frac{d{{| v |}^{2}}}{dt}=0\Rightarrow 2f^{2}t-2uf\cos \alpha =0 $
$ \Rightarrow t=\frac{u\cos \alpha }{f} $ Also $ \frac{d^{2}{{| v |}^{2}}}{dt^{2}}=2f^{2}=+ve $
$ \therefore {{| v |}^{2}} $ and hence $ | v | $ is least at the time $ \frac{u\cos \alpha }{f} $
BETA
