Starting from rest, the acceleration of a particle is a=2(t-1). The velocity of the particle at t=5s is :-

Options

A. At `t=2 s`, the position of the particle is at `x=9m`
B. Particle acceleration at `t=2 s` is `1 m//s^(2)`
C. Particle acceleration is `1//2 m//s^(2)` through out the motion
D. Particle will never go in negative direction from it's starting position

Explanation

To find the velocity of a particle at \( t = 5 \) seconds given that its acceleration is \( a = 2(t - 1) \) and it starts from rest, we can follow these steps:

### Step 1: Understand the relationship between acceleration, velocity, and time.
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]
Thus, we can express the velocity as the integral of acceleration:
\[ v = \int a \, dt + C \]

### Step 2: Substitute the expression for acceleration.
Given \( a = 2(t - 1) \), we substitute this into the integral:
\[ v = \int 2(t - 1) \, dt + C \]

### Step 3: Perform the integration.
Integrating \( 2(t - 1) \):
\[ v = 2 \left( \frac{t^2}{2} - t \right) + C = t^2 - 2t + C \]

### Step 4: Determine the constant of integration \( C \).
Since the particle starts from rest, we know that at \( t = 0 \), the velocity \( v = 0 \):
\[ v(0) = 0^2 - 2(0) + C = 0 \implies C = 0 \]
Thus, the velocity function simplifies to:
\[ v = t^2 - 2t \]

### Step 5: Calculate the velocity at \( t = 5 \) seconds.
Now, we substitute \( t = 5 \) into the velocity equation:
\[ v(5) = 5^2 - 2(5) = 25 - 10 = 15 \, \text{m/s} \]

### Conclusion:
The velocity of the particle at \( t = 5 \) seconds is \( 15 \, \text{m/s} \).

---