A particle returns to the starting point after 10s if the rate of change of velocity during the motion is constant, then its location after 7s will be same as that after
A
1s
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B
2s
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C
3s
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D
4s
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Solution
The correct option is C3s As the particle returns to its starting position after 10s, we can say 0=u×(10)−12×a×(10)2 (By second equation of motion) ⇒ua=5 Distance at which the particle will change its direction of motion is given by 0=u−at ⇒t=ua=5s ⇒Location of the particle at (t=7s) = Location of the particle at (t=3s)