If acceleration `a(t) = 3t^(2)` and initial velocity u=0 m/s , then the velocity of the particle after time t=4 s

Options

A. 73 cm/s
B. 80 cm/s
C. 64 cm/s
D. 90 cm/s

Answer

To find the velocity of the particle after time \( t = 4 \) seconds, given the acceleration \( a(t) = 3t^2 \) and the initial velocity \( u = 0 \) m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Relationship Between Acceleration and Velocity**:
The acceleration \( a(t) \) is the derivative of velocity \( v(t) \) with respect to time \( t \):
\[ a(t) = \frac{dv}{dt} \]
Given \( a(t) = 3t^2 \), we can write:
\[ \frac{dv}{dt} = 3t^2 \]

2. **Separate Variables**:
We can rearrange the equation to separate the variables \( v \) and \( t \):
\[ dv = 3t^2 dt \]

3. **Integrate Both Sides**:
We will integrate both sides. The left side will be integrated with respect to \( v \) from \( 0 \) to \( v \), and the right side will be integrated with respect to \( t \) from \( 0 \) to \( 4 \):
\[ \int_0^v dv = \int_0^4 3t^2 dt \]

4. **Calculate the Right Side Integral**:
The integral of \( 3t^2 \) with respect to \( t \) is:
\[ \int 3t^2 dt = t^3 \]
Evaluating this from \( 0 \) to \( 4 \):
\[ \left[ t^3 \right]_0^4 = 4^3 - 0^3 = 64 - 0 = 64 \]

5. **Set Up the Equation**:
Now, substituting back into our equation:
\[ v - 0 = 64 \]
Therefore, the final velocity \( v \) is:
\[ v = 64 \text{ m/s} \]

### Final Answer:
The velocity of the particle after \( t = 4 \) seconds is \( 64 \) m/s.

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