The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity `v_0`. The distance travelled by the particle in time t will be

Options

A. `v_0 t + 1/3 bt^2`
B. `v_0t + 1/3 bt^3`
C. `v_0 t + 1/6 bt^3`
D. `v_0 t + 1/2 bt^2`

Answer

C

Explanation

` a= bt`
`rArr (dv)/(dt) = bt rArr = bt dt rArr v = (bt^2)/(2) + C`
`rArr (dv)/(dt) = bt rArr dv = bt dt rArr v = (bt^2)/(2) + C`
At t = 0, `v = v_0 rArr C = v_0`
`therefore v = (bt^2)/(2) + v_0`
`(dx)/(dt) + 1/2 bt^2 + v_0`
`rArr x = 1/2 (bt^3)/(3) +v_0 t + C`
`t = 0, x = 0`
`C.= 0 `
`x = 1/6 bt^2 + v_0 t` .