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 \[2s=18a-8a=10a\] \[\therefore \] \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=2a:6a:10a\] \[{{d}_{1}}:{{d}_{2}}:{{d}_{3}}=1:3:5\]You need to login to perform this action.
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