question_answer

27.          A particle starts from rest and experiences constant acceleration for 6 s. If it travels a distance \[{{d}_{1}}\] 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)                  \[{{d}_{1}}:\text{ }{{d}_{2}}:\text{ }{{d}_{3}}=\text{ }1:1:1\]

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

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

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

Correct Answer: C

Solution :

                Let particle start from O 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: \[t=2s,\text{ }u=0\]                 \[{{d}_{1}}=\frac{1}{2}a{{(2)}^{2}}=2a\] For OB: \[t=4s,\text{ }u=0\] \[\therefore \]  \[{{d}_{2}}=\frac{1}{2}a{{(4)}^{2}}=8a\] For OC: \[u=0,\text{ }t=6\text{ }s\] \[\therefore \] \[S=\frac{1}{2}a{{(6)}^{2}}=18a\] Distance in last \[2s=18a-8a=10a\] \[\therefore \] \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=2a:6a:10a\] \[{{d}_{1}}:{{d}_{2}}:d{{ & }_{3}}=1:3:5\]