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)/(6)bt^(3)`
B. `v_(0)t+(1)/(3)bt^(3)`
C. `v_(0)t+(1)/(3)bt^(2)`
D. `v_(0)t+(1)/(2)bt^(2)`

Explanation

To solve the problem, we need to find the distance traveled by a particle whose acceleration increases linearly with time. We will follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and displacement.
The acceleration \( a(t) \) is given as \( a(t) = bt \), where \( b \) is a constant. We know that acceleration is the rate of change of velocity with respect to time.

### Step 2: Relate acceleration to velocity.
Using the definition of acceleration:
\[ a = \frac{dv}{dt} = bt \]
We can rearrange this to express \( dv \):
\[ dv = bt \, dt \]

### Step 3: Integrate to find velocity.
Now we will integrate both sides. The initial velocity is \( v_0 \) at \( t = 0 \), and we want to find the velocity \( v \) at time \( t \):
\[ \int_{v_0}^{v} dv = \int_{0}^{t} bt \, dt \]
The left side integrates to:
\[ v - v_0 \]
The right side integrates to:
\[ \frac{b}{2} t^2 \]
Thus, we have:
\[ v - v_0 = \frac{b}{2} t^2 \]
Rearranging gives us the final velocity:
\[ v = v_0 + \frac{b}{2} t^2 \]

### Step 4: Relate velocity to displacement.
We know that velocity \( v \) is the rate of change of displacement \( s \):
\[ v = \frac{ds}{dt} \]
Substituting the expression for \( v \):
\[ \frac{ds}{dt} = v_0 + \frac{b}{2} t^2 \]

### Step 5: Integrate to find displacement.
Now we will integrate to find the displacement \( s \):
\[ \int_{0}^{s} ds = \int_{0}^{t} \left( v_0 + \frac{b}{2} t^2 \right) dt \]
The left side integrates to \( s \), and the right side can be integrated as follows:
\[ s = v_0 t + \frac{b}{2} \int_{0}^{t} t^2 \, dt \]
The integral of \( t^2 \) is:
\[ \frac{t^3}{3} \]
Thus, we have:
\[ s = v_0 t + \frac{b}{2} \cdot \frac{t^3}{3} \]
This simplifies to:
\[ s = v_0 t + \frac{bt^3}{6} \]

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