The position of a particle is given by
`x=2(t-t^(2))`
where `t` is expressed in seconds and `x` is in metre.
The total distance travelled by the paticle between `t=0` to `t=1s` is

Options

A. `0m`
B. `1m`
C. `2m`
D. `1/2m`

Answer

To find the total distance travelled by the particle between \( t = 0 \) seconds and \( t = 1 \) second, we start with the position function given by:

\[ x(t) = 2(t - t^2) \]

### Step 1: Determine the position at \( t = 0 \) seconds.
Substituting \( t = 0 \) into the position function:

\[ x(0) = 2(0 - 0^2) = 2(0) = 0 \text{ meters} \]

### Step 2: Determine the position at \( t = 1 \) second.
Substituting \( t = 1 \) into the position function:

\[ x(1) = 2(1 - 1^2) = 2(1 - 1) = 2(0) = 0 \text{ meters} \]

### Step 3: Find the position at \( t = \frac{1}{2} \) seconds to check for maximum displacement.
Substituting \( t = \frac{1}{2} \) into the position function:

\[ x\left(\frac{1}{2}\right) = 2\left(\frac{1}{2} - \left(\frac{1}{2}\right)^2\right) = 2\left(\frac{1}{2} - \frac{1}{4}\right) = 2\left(\frac{1}{4}\right) = \frac{1}{2} \text{ meters} \]

### Step 4: Analyze the motion of the particle.
From \( t = 0 \) to \( t = \frac{1}{2} \) seconds, the particle moves from \( 0 \) meters to \( \frac{1}{2} \) meters.

From \( t = \frac{1}{2} \) seconds to \( t = 1 \) second, the particle moves back from \( \frac{1}{2} \) meters to \( 0 \) meters.

### Step 5: Calculate the total distance travelled.
The total distance travelled is the sum of the distances in both segments of the motion:

1. Distance from \( 0 \) to \( \frac{1}{2} \) meters: \( \frac{1}{2} \) meters
2. Distance from \( \frac{1}{2} \) meters back to \( 0 \) meters: \( \frac{1}{2} \) meters

Thus, the total distance travelled is:

\[ \text{Total Distance} = \frac{1}{2} + \frac{1}{2} = 1 \text{ meter} \]

### Final Answer:
The total distance travelled by the particle between \( t = 0 \) seconds and \( t = 1 \) second is \( 1 \) meter.

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