The acceleration of 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_0 t + (1)/(3) bt^3`
C. `v_0 t + (1)/(6) bt^2`
D. `v_0 t + (1)/(2) bt^2`

Explanation

To solve this problem, we need to find the distance traveled by a particle whose acceleration increases linearly with time. Let's break down the problem step by step.

### Step 1: Express the acceleration
Given that the acceleration \( a(t) \) is increasing linearly with time \( t \):
\[ a(t) = bt \]

### Step 2: Integrate to find the velocity
Acceleration \( a(t) \) is the derivative of velocity \( v(t) \) with respect to time:
\[ \frac{dv}{dt} = bt \]

To find the velocity, integrate both sides with respect to \( t \):
\[ \int dv = \int bt \, dt \]

The limits of integration for \( v \) are from \( v_0 \) (initial velocity at \( t = 0 \)) to \( v \) (velocity at time \( t \)), and for \( t \) from 0 to \( t \):
\[ v - v_0 = \int_0^t bt \, dt \]

Perform the integration:
\[ v - v_0 = b \int_0^t t \, dt \]
\[ v - v_0 = b \left[ \frac{t^2}{2} \right]_0^t \]
\[ v - v_0 = b \left( \frac{t^2}{2} - \frac{0^2}{2} \right) \]
\[ v - v_0 = \frac{bt^2}{2} \]

Thus, the velocity as a function of time is:
\[ v = v_0 + \frac{bt^2}{2} \]

### Step 3: Integrate to find the position
Velocity \( v(t) \) is the derivative of position \( x(t) \) with respect to time:
\[ \frac{dx}{dt} = v_0 + \frac{bt^2}{2} \]

To find the position, integrate both sides with respect to \( t \):
\[ \int dx = \int \left( v_0 + \frac{bt^2}{2} \right) dt \]

The limits of integration for \( x \) are from 0 (initial position at \( t = 0 \)) to \( x \) (position at time \( t \)), and for \( t \) from 0 to \( t \):
\[ x - 0 = \int_0^t \left( v_0 + \frac{bt^2}{2} \right) dt \]

Perform the integration:
\[ x = \int_0^t v_0 \, dt + \int_0^t \frac{bt^2}{2} \, dt \]
\[ x = v_0 \left[ t \right]_0^t + \frac{b}{2} \int_0^t t^2 \, dt \]
\[ x = v_0 \left( t - 0 \right) + \frac{b}{2} \left[ \frac{t^3}{3} \right]_0^t \]
\[ x = v_0 t + \frac{b}{2} \left( \frac{t^3}{3} - \frac{0^3}{3} \right) \]
\[ x = v_0 t + \frac{b t^3}{6} \]

Thus, the distance traveled by the particle in time \( t \) is:
\[ x(t) = v_0 t + \frac{b t^3}{6} \]

### Final Answer
The distance traveled by the particle in time \( t \) is:
\[ x(t) = v_0 t + \frac{b t^3}{6} \]