A starts from rest, with uniform acceleration a. The acceleration of the body as function of time t is given by the equation a = pt, where p is a constant, then the displacement of the particle in the time interval t = 0 to `t=t_(1)` will be

Options

A. `(1)/(2)pt_(1)^(3)`
B. `(1)/(3)pt_(1)^(2)`
C. `(1)/(2)pt_(1)^(2)`
D. `(1)/(2)pt_(1)^(3)`

Explanation

To solve the problem step by step, we need to find the displacement of a particle that starts from rest and has an acceleration that varies with time. The acceleration is given by the equation \( a = pt \), where \( p \) is a constant.

### Step 1: Understand the relationship between acceleration, velocity, and displacement
We know that:
- Acceleration \( a = \frac{dv}{dt} \)
- Velocity \( v = \frac{dx}{dt} \)

### Step 2: Express acceleration in terms of time
Given that \( a = pt \), we can write:
\[ \frac{dv}{dt} = pt \]

### Step 3: Integrate to find velocity
To find the velocity \( v \), we integrate the acceleration with respect to time \( t \):
\[ dv = pt \, dt \]
Integrating both sides:
\[ v = \int pt \, dt = \frac{p}{2} t^2 + C \]
Since the particle starts from rest, the initial velocity \( v(0) = 0 \), which means \( C = 0 \). Thus, we have:
\[ v = \frac{p}{2} t^2 \]

### Step 4: Express velocity in terms of time
Now we have the expression for velocity:
\[ v = \frac{p}{2} t^2 \]

### Step 5: Integrate to find displacement
Next, we need to find the displacement \( x \) by integrating the velocity:
\[ dx = v \, dt = \frac{p}{2} t^2 \, dt \]
Integrating both sides from \( t = 0 \) to \( t = t_1 \):
\[ x = \int_0^{t_1} \frac{p}{2} t^2 \, dt \]
Calculating the integral:
\[ x = \frac{p}{2} \cdot \left[ \frac{t^3}{3} \right]_0^{t_1} = \frac{p}{2} \cdot \frac{t_1^3}{3} = \frac{p t_1^3}{6} \]

### Final Result
Thus, the displacement of the particle in the time interval from \( t = 0 \) to \( t = t_1 \) is:
\[ x = \frac{p t_1^3}{6} \]

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