A particle is moving on a straight line. Its velocity at time `t` is `(8-2t)m//s`. What is the total distance covered from `t=0` to `t=6s`?

Options

A. `12m`
B. `16m`
C. `18m`
D. `20m`

Answer

To find the total distance covered by the particle from \( t = 0 \) to \( t = 6 \) seconds, we start with the given velocity function:

\[ v(t) = 8 - 2t \quad \text{(in m/s)} \]

### Step 1: Determine when the particle changes direction
First, we need to find out if the particle changes direction during the interval from \( t = 0 \) to \( t = 6 \) seconds. This occurs when the velocity \( v(t) \) becomes zero.

Set the velocity to zero:

\[ 8 - 2t = 0 \]

Solving for \( t \):

\[ 2t = 8 \implies t = 4 \text{ seconds} \]

### Step 2: Calculate the distance for each segment
Now we will calculate the distance covered in two segments: from \( t = 0 \) to \( t = 4 \) seconds (when the particle is moving in one direction) and from \( t = 4 \) to \( t = 6 \) seconds (when the particle is moving in the opposite direction).

#### Segment 1: From \( t = 0 \) to \( t = 4 \)
We need to find the displacement during this time. We can integrate the velocity function:

\[ \text{Displacement} = \int_{0}^{4} (8 - 2t) \, dt \]

Calculating the integral:

\[ = \int_{0}^{4} 8 \, dt - \int_{0}^{4} 2t \, dt \]

Calculating each integral:

\[ = 8t \bigg|_{0}^{4} - 2 \cdot \frac{t^2}{2} \bigg|_{0}^{4} \]

Evaluating the limits:

\[ = 8(4) - (4^2) = 32 - 16 = 16 \text{ meters} \]

#### Segment 2: From \( t = 4 \) to \( t = 6 \)
Now we calculate the displacement from \( t = 4 \) to \( t = 6 \):

\[ \text{Displacement} = \int_{4}^{6} (8 - 2t) \, dt \]

Calculating the integral:

\[ = \int_{4}^{6} 8 \, dt - \int_{4}^{6} 2t \, dt \]

Calculating each integral:

\[ = 8t \bigg|_{4}^{6} - 2 \cdot \frac{t^2}{2} \bigg|_{4}^{6} \]

Evaluating the limits:

\[ = 8(6) - 8(4) - (6^2 - 4^2) = 48 - 32 - (36 - 16) = 16 - 20 = -4 \text{ meters} \]

Since distance cannot be negative, we take the absolute value:

\[ \text{Distance} = 4 \text{ meters} \]

### Step 3: Total distance covered
Now, we add the distances from both segments:

\[ \text{Total Distance} = 16 \text{ meters} + 4 \text{ meters} = 20 \text{ meters} \]

Thus, the total distance covered by the particle from \( t = 0 \) to \( t = 6 \) seconds is:

\[ \boxed{20 \text{ meters}} \]