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_0t+(1)/(3)bt^2`
B. `v_0t+(1)/(3)bt^3`
C. `v_0t+(1)/(6)bt^3`
D. `v_0t+(1)/(2)bt^2`

Answer

C

Explanation

`(dv)/(dt)=btimpliesdv=btdtimpliesv=(bt^2)/(2)+K_1`
At `t=0`,`v=0impliesK-1=v_0`
we get `v=(1)/(2)bt^2+v_0`
Again `(dx)/(dt)=(1)/(2)bt^2+v_0impliesx=(1)/(2)(bt^2)/(3)+v_0t+K-2`
At `t=0`,`x=0impliesK-2=0`
`x=(1)/(6)bt^3+v_0t`