Acceleration time graph of a particle is shown work done by all the force acting on the particle of mass `m` in line internal`t_(1)` and `t_(2)` while `a_(1)` is the acceleration at time `t_(1)` given by `t = 0` particle was at rest

Options

A. `(ma_(1)^(2))/(4t_(1)) (t_(2)^(3)- t_(1)^(2))`
B. `(ma_(1)^(2))/(8t_(1)^2) (t_(2)^(4)- t_(1)^(4))`
C. `(ma_(1)^(2))/(4t_(1)^(2))(t_(2)^(4)- t_(1)^(4))`
D. `(ma_(1))/(4t_(1)) (t_(2)^(2)- t_(1)^(2))`

Answer

B

Explanation

`W = K_(f) - K_(i) = (1)/(2)(mv_(1)^(2) - v_(1)^(2))`
Area under acceleration - time graph is change velocity
so, `W = (1)/(2)m[((1)/(2) a_(2)t_(2))^(2) - ((1)/(2) a_(1)t_(1))^(2))]`
`= (1)/(8)m [(t_(2)^(4))/(t_(1)^(2)) a_(1)^(2) - a_(1)^(2) t_(1)^(2)]` (since `a_(2) = (t_(2))/(t_(1))a_(1)`)
`= (ma_(1)^(2))/(8t_(1)^(2)) (t_(2)^(4) - t_(1)^(4))`.