Starting from rest a particle moves in a straight line with acceleration
`a = (25 - t^(2))^(1//2) m//s^(2) " for " 0 le t le 5s`
`a = (3pi)/(8) m//s^(2) " for " t gt 5s`
The velocity of particle at `t = 7 s` is:

Options

A. 11m/s
B. 22m/s
C. 33m/s
D. 44m/s

Answer

B

Explanation

`V_(7) = V_(5) + (3pi)/(8) xx 2` To calculate `V_(5)` i.e., velocity at end of 5 sec.
`a= (25-t^(2))^(1//2) rArr (dv)/(dt) = (25-t^(2))^(1//2)`
`int dv= int 5 (25-25 sin^(2) theta)^(1//2) cos theta d theta {(t=5 sin theta),((dt)/(d theta) = 5 cos theta),(t0"," theta= 0),(t =5"," theta= (pi)/(2)):}`
`= underset(0)overset(pi//2)int 5 xx 4 cos^(2) theta d theta = (25pi)/(4)`
`V_(7) = (25pi)/(4) + (3pi)/(4) = (28pi)/(4) = 7pi = 22 m//s`