Velocity time equation of a particle moving in a straight line is `v=2t-4` for `tle2s` and `v=4-2t` for `tgt2`.The distance travelled by the particle in the time interval from `t=0` to `t=4s ` is (Here `t` is in second and `v` in m/s)

Options

A. `12m`
B. `16m`
C. `4m`
D. `8m`

Answer

To solve the problem, we need to calculate the distance traveled by the particle in the time interval from \( t = 0 \) to \( t = 4 \) seconds using the given velocity equations.

### Step 1: Identify the time intervals and corresponding velocity equations
The velocity equations are:
- For \( t \leq 2 \) seconds: \( v = 2t - 4 \)
- For \( t > 2 \) seconds: \( v = 4 - 2t \)

### Step 2: Calculate the distance from \( t = 0 \) to \( t = 2 \) seconds
We will integrate the velocity function from \( t = 0 \) to \( t = 2 \):
\[ \text{Distance}_1 = \int_{0}^{2} (2t - 4) \, dt \]

Calculating the integral:
\[ \int (2t - 4) \, dt = t^2 - 4t \]
Now, evaluate it from 0 to 2:
\[ \text{Distance}_1 = \left[ t^2 - 4t \right]_{0}^{2} = (2^2 - 4 \cdot 2) - (0^2 - 4 \cdot 0) = (4 - 8) - 0 = -4 \]
Since distance cannot be negative, we take the absolute value:
\[ \text{Distance}_1 = | -4 | = 4 \text{ meters} \]

### Step 3: Calculate the distance from \( t = 2 \) to \( t = 4 \) seconds
Now we will integrate the velocity function from \( t = 2 \) to \( t = 4 \):
\[ \text{Distance}_2 = \int_{2}^{4} (4 - 2t) \, dt \]

Calculating the integral:
\[ \int (4 - 2t) \, dt = 4t - t^2 \]
Now, evaluate it from 2 to 4:
\[ \text{Distance}_2 = \left[ 4t - t^2 \right]_{2}^{4} = (4 \cdot 4 - 4^2) - (4 \cdot 2 - 2^2) = (16 - 16) - (8 - 4) = 0 - 4 = -4 \]
Again, taking the absolute value:
\[ \text{Distance}_2 = | -4 | = 4 \text{ meters} \]

### Step 4: Calculate the total distance traveled
Now, we add the distances from both intervals:
\[ \text{Total Distance} = \text{Distance}_1 + \text{Distance}_2 = 4 + 4 = 8 \text{ meters} \]

### Final Answer
The distance traveled by the particle from \( t = 0 \) to \( t = 4 \) seconds is **8 meters**.

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