In previous problem, average speed during the motion of the particle is.

Options

A. `8 m//s`
B. zero
C. `4 m//s`
D. `16 m//s`

Answer

A

Explanation

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`.