The displacement of a particle moving in a straight line is described by the relation `s=6+12t-2t^(2)`. Here `s` is in metre and `t` in second. The distance covered by the particle in first `5s` is

Options

A. `20m`
B. `32m`
C. `24m`
D. `26m`

Answer

To find the distance covered by the particle in the first 5 seconds, we start with the given displacement equation:

\[ s = 6 + 12t - 2t^2 \]

### Step 1: Find the velocity of the particle
To determine if the particle changes direction, we first need to find the velocity by differentiating the displacement equation with respect to time \( t \):

\[ v = \frac{ds}{dt} = 12 - 4t \]

### Step 2: Determine when the velocity is zero
Next, we set the velocity equation to zero to find the time when the particle changes direction:

\[ 12 - 4t = 0 \]

Solving for \( t \):

\[ 4t = 12 \]
\[ t = 3 \, \text{seconds} \]

### Step 3: Calculate displacement from \( t = 0 \) to \( t = 3 \)
Now we calculate the displacement from \( t = 0 \) to \( t = 3 \):

\[ s(3) = 6 + 12(3) - 2(3^2) \]
\[ s(3) = 6 + 36 - 18 \]
\[ s(3) = 24 \, \text{meters} \]

At \( t = 0 \):

\[ s(0) = 6 + 12(0) - 2(0^2) = 6 \, \text{meters} \]

So, the displacement from \( t = 0 \) to \( t = 3 \) is:

\[ s_1 = s(3) - s(0) = 24 - 6 = 18 \, \text{meters} \]

### Step 4: Calculate displacement from \( t = 3 \) to \( t = 5 \)
Next, we calculate the displacement from \( t = 3 \) to \( t = 5 \):

\[ s(5) = 6 + 12(5) - 2(5^2) \]
\[ s(5) = 6 + 60 - 50 \]
\[ s(5) = 16 \, \text{meters} \]

Now, the displacement from \( t = 3 \) to \( t = 5 \):

\[ s_2 = s(5) - s(3) = 16 - 24 = -8 \, \text{meters} \]

### Step 5: Calculate the total distance covered
The total distance covered is the sum of the magnitudes of the displacements:

\[ \text{Total Distance} = |s_1| + |s_2| \]
\[ \text{Total Distance} = 18 + 8 = 26 \, \text{meters} \]

### Final Answer
The distance covered by the particle in the first 5 seconds is:

\[ \boxed{26 \, \text{meters}} \]