Kinematics Question 160
Question: A particle moves with uniform acceleration along a straight line AB. it’s velocities at A and B are 2 m/s and 14 m/s, respectively. M is the mid-point of AB. The particle takes $ t _{1} $ seconds to go from A to M and $ t _{2} $ seconds to go from M to B. Then $ t _{2}/t _{1} $ it’s
Options:
A) 1 : 1
B) 2 : 1
C) 1:2
D) 3:1
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ V^{2}=2^{2}+\frac{2as}{2} $
$ \Rightarrow v^{2}-4=as $
$ 14^{2}-v^{2}=\frac{2as}{2} $
$ \Rightarrow 196-v^{2}=as $ From (i) and (ii) $ v^{2}-4=196-v^{2}\Rightarrow v=10m/s $ Now $ t _{1}=\frac{v-2}{a}=\frac{10-2}{a}=\frac{8}{a} $
$ t _{2}=\frac{14-v}{a}=\frac{14-10}{a}=\frac{4}{a}\Rightarrow \frac{t _{2}}{t^{1}}=\frac{4/a}{8/a}=\frac{1}{2} $
