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 given information and apply the appropriate physics concepts.

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

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

### Step 3: Integrate to find velocity
To find the velocity \( v \), we need to 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 find 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:
\[ v = \frac{ds}{dt} \]
Substituting the expression for velocity:
\[ 3t^2 = \frac{ds}{dt} \]

### Step 5: Integrate to find distance
Now we integrate to find the distance \( s \):
\[ \int ds = \int 3t^2 \, dt \]
This gives:
\[ s = t^3 + C' \]
Since the particle starts from the origin (initial position \( s = 0 \) when \( t = 0 \)), we can find the constant \( C' \):
\[ 0 = (0)^3 + C' \implies C' = 0 \]
Thus, the distance function is:
\[ s = t^3 \]

### Step 6: Final expression for distance
The distance traveled by the particle in time \( t \) is:
\[ s = t^3 \]

### Conclusion
The distance traveled by the particle in time \( t \) is \( t^3 \).

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