The acceleration `a` of a particle in `m/s^2` is give by `a=18t^2+36t+120`. If the particle starts from rest, then velocity at end of 1 s is

Options

A. 122 m/s
B. 140 m/s
C. 144 m/s
D. 120 m/s

Answer

To find the velocity of the particle at the end of 1 second, we need to integrate the acceleration function with respect to time. Let's break down the steps:

### Step-by-Step Solution

1. **Understand the Given Information:**
- The acceleration \( a \) is given by the equation:
\[ a = 18t^2 + 36t + 120 \]
- The particle starts from rest, which means the initial velocity \( u = 0 \).

2. **Set Up the Integral for Velocity:**
- The change in velocity \( \Delta v \) can be found by integrating the acceleration over the time interval from \( t = 0 \) to \( t = 1 \):
\[ v = \int_{0}^{1} a \, dt \]

3. **Substitute the Acceleration Function:**
- Substitute the expression for acceleration into the integral:
\[ v = \int_{0}^{1} (18t^2 + 36t + 120) \, dt \]

4. **Perform the Integration:**
- Integrate each term separately:
- For \( 18t^2 \):
\[ \int 18t^2 \, dt = 18 \cdot \frac{t^3}{3} = 6t^3 \]
- For \( 36t \):
\[ \int 36t \, dt = 36 \cdot \frac{t^2}{2} = 18t^2 \]
- For \( 120 \):
\[ \int 120 \, dt = 120t \]
- Combining these results, we have:
\[ v = \left[ 6t^3 + 18t^2 + 120t \right]_{0}^{1} \]

5. **Evaluate the Integral from 0 to 1:**
- Substitute \( t = 1 \):
\[ v(1) = 6(1)^3 + 18(1)^2 + 120(1) = 6 + 18 + 120 = 144 \]
- Substitute \( t = 0 \):
\[ v(0) = 6(0)^3 + 18(0)^2 + 120(0) = 0 \]
- Therefore, the change in velocity is:
\[ v = 144 - 0 = 144 \, \text{m/s} \]

### Final Answer:
The velocity of the particle at the end of 1 second is \( 144 \, \text{m/s} \).

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