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

Options

A. 15 m/s
B. 25 m/s
C. 5 m/s
D. None of these

Explanation

To solve the problem, we need 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.

### Step-by-Step Solution:

1. **Understand the relationship between acceleration and velocity**:
The acceleration \( a \) is the derivative of velocity \( v \) with respect to time \( t \):
\[ a = \frac{dv}{dt} \]

2. **Set up the equation for acceleration**:
Given the acceleration:
\[ a = 2(t - 1) \]

3. **Integrate the acceleration to find velocity**:
To find the velocity \( v \), we integrate the acceleration with respect to time:
\[ v = \int a \, dt = \int 2(t - 1) \, dt \]

4. **Perform the integration**:
\[ v = \int (2t - 2) \, dt = 2 \int t \, dt - 2 \int 1 \, dt \]
\[ v = 2 \left( \frac{t^2}{2} \right) - 2t + C \]
\[ v = t^2 - 2t + C \]

5. **Determine the constant of integration \( C \)**:
Since the particle starts from rest, we know that at \( t = 0 \), the velocity \( v = 0 \):
\[ 0 = (0)^2 - 2(0) + C \implies C = 0 \]
Therefore, the equation for velocity simplifies to:
\[ v = t^2 - 2t \]

6. **Calculate the velocity at \( t = 5 \) seconds**:
Substitute \( t = 5 \) into the velocity equation:
\[ v(5) = 5^2 - 2(5) \]
\[ v(5) = 25 - 10 = 15 \, \text{m/s} \]

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

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