The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v₀. The distance travelled by the particle in time t will be

The correct option is v₀t+(1/6)bt³.

Let's find the distance travelled by the particle step-by-step using calculus.

Step 1: Express Acceleration as a function of time
The acceleration $a$ of the particle is increasing linearly with time $t$ as $bt$. Therefore, we can write:
$a=\frac{dv}{dt}=bt$
where $v$ is the velocity of the particle at any time $t$.

Step 2: Find Velocity by integration
Rearranging the terms, we get:
$dv=btdt$
Integrating both sides, taking the limits from $t=0$ (where velocity is $v_0$) to time $t$ (where velocity is $v$):
${\int }_{v_0}^{v}dv={\int }_0^{t}btdt$
${\left[v\right]}_{v_0}^v=b{\left[\frac{t^2}{2}\right]}_0^t$
$v-v_0=\frac{1}{2}b{t}^2$
$v=v_0+\frac{1}{2}b{t}^2$

Step 3: Find Distance by integration
Velocity $v$ is defined as the rate of change of position $s$ with respect to time:
$v=\frac{ds}{dt}=v_0+\frac{1}{2}b{t}^2$
Rearranging the terms to integrate:
$ds=(v_0+\frac{1}{2}b{t}^2)dt$
Since the particle starts from the origin, at $t=0$, the position $s=0$. Integrating from $t=0$ to time $t$:
${\int }_0^{s}ds={\int }_0^t(v_0+\frac{1}{2}b{t}^2)dt$
$s={[v_{0}t+\frac{1}{2}b\frac{t^3}{3}]}_0^t$
$s=v_{0}t+\frac{1}{6}b{t}^3$

Thus, the distance travelled by the particle in time $t$ is $v_{0}t+\frac{1}{6}b{t}^3$.