If velocity of a moving particle in is `v= a + g t + ft^2` (a,g,f are constants). At t=0 body is at origin. Find displacement after t=1s.

Options

A. `a+g+f`
B. `g+2f`
C. `a+(g/2)+(f/3)`
D. `(a/2) + (g/3) + (f/4)`

Answer

To find the displacement of a moving particle after \( t = 1 \) second, given the velocity function \( v(t) = a + gt + ft^2 \), we will follow these steps:

### Step-by-Step Solution:

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

2. **Express the velocity function**:
Given the velocity function:
\[ v(t) = a + gt + ft^2 \]

3. **Set up the integral for displacement**:
To find the displacement \( s \) after \( t = 1 \) second, we need to integrate the velocity function from \( t = 0 \) to \( t = 1 \):
\[ s = \int_{0}^{1} v(t) \, dt = \int_{0}^{1} (a + gt + ft^2) \, dt \]

4. **Integrate the velocity function**:
We can break down the integral:
\[ s = \int_{0}^{1} a \, dt + \int_{0}^{1} gt \, dt + \int_{0}^{1} ft^2 \, dt \]
- The first integral:
\[ \int_{0}^{1} a \, dt = a \cdot t \bigg|_{0}^{1} = a \cdot 1 - a \cdot 0 = a \]
- The second integral:
\[ \int_{0}^{1} gt \, dt = g \cdot \frac{t^2}{2} \bigg|_{0}^{1} = g \cdot \frac{1^2}{2} - g \cdot \frac{0^2}{2} = \frac{g}{2} \]
- The third integral:
\[ \int_{0}^{1} ft^2 \, dt = f \cdot \frac{t^3}{3} \bigg|_{0}^{1} = f \cdot \frac{1^3}{3} - f \cdot \frac{0^3}{3} = \frac{f}{3} \]

5. **Combine the results**:
Now, we can combine all the results from the integrals:
\[ s = a + \frac{g}{2} + \frac{f}{3} \]

6. **Final expression for displacement**:
Therefore, the displacement after \( t = 1 \) second is:
\[ s = a + \frac{g}{2} + \frac{f}{3} \]