The relation between the acceleration and time for an object is given below. Calculate the velocity with which the object is moving at t = 1 s.
`a=3t-4t^(2)`

Explanation

To solve the problem, we need to find the velocity of the object at \( t = 1 \) second given the acceleration function \( a(t) = 3t - 4t^2 \).

### Step-by-Step Solution:

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

2. **Substitute the given acceleration function**:
We can substitute the given acceleration function into the equation:
\[ \frac{dv}{dt} = 3t - 4t^2 \]

3. **Rearrange the equation for integration**:
Rearranging gives us:
\[ dv = (3t - 4t^2) dt \]

4. **Integrate both sides**:
Now, we integrate both sides. The left side integrates to \( v \), and the right side needs to be integrated with respect to \( t \):
\[ \int dv = \int (3t - 4t^2) dt \]

5. **Perform the integration**:
The integral of the left side is:
\[ v = \int (3t - 4t^2) dt = \frac{3}{2}t^2 - \frac{4}{3}t^3 + C \]
where \( C \) is the constant of integration.

6. **Evaluate the integral at \( t = 1 \) second**:
We need to find the velocity at \( t = 1 \) second. We substitute \( t = 1 \) into the integrated equation:
\[ v(1) = \frac{3}{2}(1^2) - \frac{4}{3}(1^3) + C \]
This simplifies to:
\[ v(1) = \frac{3}{2} - \frac{4}{3} + C \]

7. **Find a value for the constant \( C \)**:
Since we don't have initial conditions provided (like initial velocity), we can assume \( C = 0 \) for simplicity. Thus:
\[ v(1) = \frac{3}{2} - \frac{4}{3} \]

8. **Calculate the final value**:
To combine the fractions, we find a common denominator (which is 6):
\[ v(1) = \frac{9}{6} - \frac{8}{6} = \frac{1}{6} \]

9. **Convert to decimal**:
The final velocity at \( t = 1 \) second is:
\[ v(1) = 0.1667 \text{ m/s} \quad (\text{approximately}) \]

### Final Answer:
The velocity of the object at \( t = 1 \) second is approximately \( 0.1667 \, \text{m/s} \).