The velocity –position graph of a particle moving along a straight line is shown. Then acceleration of the particle at s = 15 m is :

Options

A. ` (2)/(3) ms^(-2) `
B. ` -(2)/(3) ms^(-2) `
C. ` (20)/(3)ms^(-2)`
D. ` -(20)/( 3)ms^(-2)`

Answer

D

Explanation

` V(s) =20 -(2s)/(3) rArr (15)=10 m//s(dv)/(ds) =` slope of v-s graph ` =-(2)/(3)`
`a(S) =(dv)/(ds) =(v) =((dv)/(ds)) =(10 ) ((-2)/(3)) =(-20 )/(3) m//s^(2) `