The acceleratin of a particle increases linearly with time t as 6t. If the initial velocity of the particles is zero and the particle starts from the origin, then the distance traveled by the particle in time t will be

Options

A. t
B. `t ^(2)`
C. `t ^(3)`
D. `t ^(4)`

Answer

To solve the problem step by step, we will follow the principles of kinematics, specifically focusing on the relationship between acceleration, velocity, and distance.

### Step 1: Understand the given information
We are given that the acceleration \( a \) of the particle varies with time \( t \) as:
\[ a = 6t \]
The initial velocity \( u \) is zero, and the particle starts from the origin (initial position \( x_0 = 0 \)).

### Step 2: Relate acceleration to velocity
Acceleration is defined as the rate of change of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
Substituting the expression for acceleration:
\[ \frac{dv}{dt} = 6t \]

### Step 3: Integrate to find velocity
To find the velocity \( v \), we integrate both sides with respect to time \( t \):
\[ \int dv = \int 6t \, dt \]
This gives:
\[ v = 3t^2 + C \]
Since the initial velocity \( u = 0 \) when \( t = 0 \), we can determine the constant \( C \):
\[ 0 = 3(0)^2 + C \implies C = 0 \]
Thus, the velocity function is:
\[ v = 3t^2 \]

### Step 4: Relate velocity to distance
Velocity is also defined as the rate of change of distance with respect to time:
\[ v = \frac{dx}{dt} \]
Substituting the expression for velocity:
\[ \frac{dx}{dt} = 3t^2 \]

### Step 5: Integrate to find distance
To find the distance \( x \), we integrate both sides with respect to time \( t \):
\[ \int dx = \int 3t^2 \, dt \]
This gives:
\[ x = t^3 + C' \]
Since the particle starts from the origin (initial position \( x_0 = 0 \) when \( t = 0 \)), we can determine the constant \( C' \):
\[ 0 = (0)^3 + C' \implies C' = 0 \]
Thus, the distance function is:
\[ x = t^3 \]

### Step 6: Conclusion
The distance traveled by the particle in time \( t \) is:
\[ \boxed{t^3} \]