question_answer

A particle moves with uniform acceleration along a straight line AB. Its 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  \[{{t}_{2}}/{{t}_{1}}\] is

A) 1 : 1                 

B) 2 : 1

C) 1:2                   

D) 3:1

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 \[{{t}_{1}}=\frac{v-2}{a}=\frac{10-2}{a}=\frac{8}{a}\] \[{{t}_{2}}=\frac{14-v}{a}=\frac{14-10}{a}=\frac{4}{a}\Rightarrow \frac{{{t}_{2}}}{{{t}^{1}}}=\frac{4/a}{8/a}=\frac{1}{2}\]