The acceleration time graph of the motion of a particle from A to B is semicircle with radius of `sqrt(7)` . Find the velocity of particle at `t_(2)` at it `t_(1)` velocity is `3m//sec [pi = 22//7]` .

Options

A. `(15)/(2)m//sec`
B. `8m//sec`
C. `(17)/(2)m//sec`
D. None of these

Answer

C

Explanation

The acceleration …………..
we known that change in velocity = Area under a - t curve.
`:. V_(t_(2)) - V_(t_(1))` = Area under the graph and t - axis from D to B (which is quaether circle)
`:. V_(t_(2)) - V_(t_(1)) = (pi r^(2))/(4) = (22)/(7)xx((sqrt(7))^(2))/(4) =`
`:. V_(t_(2))- 3=(11)/(2) or = (11)/(2) +3 = (17)/(2)m//sec`