A particle starts from rest with a time varying acceleration `a=(2t-4)`. Here `t` is in second and `a` in `m//s^(2)`
Particle comes to rest after a time `t=`………… second

Options

A. 1
B. 4
C. 3
D. 2

Answer

To solve the problem step by step, we will start with the given acceleration and integrate it to find the velocity, and then find the time when the particle comes to rest.

### Step 1: Write down the given acceleration
The acceleration of the particle is given by:
\[ a(t) = 2t - 4 \]

### Step 2: Relate acceleration to velocity
We know that acceleration is the derivative of velocity with respect to time:
\[ a = \frac{dv}{dt} \]
This means we can express the change in velocity in terms of acceleration:
\[ dv = a \, dt \]

### Step 3: Substitute the expression for acceleration
Substituting the expression for acceleration into the equation:
\[ dv = (2t - 4) \, dt \]

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

The left side integrates to:
\[ v = \int (2t - 4) \, dt = t^2 - 4t + C \]
where \( C \) is the constant of integration.

### Step 5: Determine the constant of integration
Since the particle starts from rest, we know that when \( t = 0 \), \( v = 0 \):
\[ 0 = (0)^2 - 4(0) + C \implies C = 0 \]
Thus, the velocity function simplifies to:
\[ v(t) = t^2 - 4t \]

### Step 6: Find when the particle comes to rest
The particle comes to rest when its velocity is zero:
\[ v(t) = 0 \implies t^2 - 4t = 0 \]
Factoring the equation gives:
\[ t(t - 4) = 0 \]
This results in two solutions:
1. \( t = 0 \) (the initial time)
2. \( t = 4 \) seconds

### Conclusion
The particle comes to rest after \( t = 4 \) seconds.