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In previous problem, average speed during the motion of the particle is.
A. `8 m//s`
B. zero
C. `4 m//s`
D. `16 m//s`

1 Answer

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Best answer
Correct Answer - A
Distance travelled
`s = (1)/(2) a_1 t_1^2 + (1)/(2) a_2 t_2^2 = (1)/(2) 4(4)^2 + (1)/(2)(2)(8)^2`
[Acceleration in second part will be half of first part because time taken is double]
`rArr s = 96 mv_(av) = (96)/(4 + 8) = 8 m//s`.

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