question_answer

A particle starts from rest and experiences constant acceleration for 6 s. If it travels a distance d-^ in the first two seconds, a distance  \[{{d}_{2}}\] in the next two seconds and a distance \[{{d}_{3}}\] in the last two seconds, then

A)  \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:1:1\]

B)  \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:2:3\]

C)  \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:3:5\]

D)  \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:5:9\]

Correct Answer: C

Solution :

Let pardcle start from 0 and travels distance \[{{d}_{1}}(OA),{{d}_{2}}(AB),{{d}_{3}}(BC).\]. From equation of motion, we have \[S=ut+\frac{1}{2}g{{t}^{2}}\] where u is initial velocity, t- time and g acceleration due to gravity. For \[OA:\,\,\,t=2s,\,\,\,\,\,u=0\] \[{{d}_{1}}=\frac{1}{2}a{{(2)}^{2}}=2a\] For \[OB:\text{ }r=4\text{ }s,\text{ }u=0\] \[\therefore \] \[{{d}_{2}}=\frac{1}{2}a{{(4)}^{2}}=8a\] For  \[OC:u=0,\text{ }t=6s\] \[\therefore \] \[S=\frac{1}{2}a{{(6)}^{2}}=18a\] Distance in last \[2s=18a-8a=10a\] \[\therefore \] \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=2a:6a:10a\] \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:3:5\]