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

Explanation

To find the distance covered by the particle in the first 5 seconds given the displacement equation \( s = 6 + 12t - 2t^2 \), we will follow these steps:

### Step 1: Determine the velocity of the particle
The velocity \( v \) can be found by differentiating the displacement \( s \) with respect to time \( t \).

\[ v = \frac{ds}{dt} = \frac{d}{dt}(6 + 12t - 2t^2) = 12 - 4t \]

### Step 2: Find when the velocity is zero
To find the time when the particle stops (velocity = 0), we set the velocity equation to zero:

\[ 12 - 4t = 0 \]

Solving for \( t \):

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

### Step 3: Calculate displacement at \( t = 0 \) and \( t = 3 \)
Now, we will calculate the displacement at \( t = 0 \) and \( t = 3 \).

- At \( t = 0 \):

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

- At \( t = 3 \):

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

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

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

### Step 5: Determine the total distance covered
Now we need to find the total distance covered in the first 5 seconds. The particle moves from \( s(0) = 6 \) to \( s(3) = 24 \) and then back to \( s(5) = 16 \).

1. Distance from \( t = 0 \) to \( t = 3 \):

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

2. Distance from \( t = 3 \) to \( t = 5 \):

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

### Step 6: Calculate the total distance
Finally, we add the distances:

\[ \text{Total Distance} = 18 + 8 = 26 \text{ meters} \]

### Conclusion
The distance covered by the particle in the first 5 seconds is **26 meters**.

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