`x` and `y` co-ordinates of a particle moving in `x-y` plane at some instant of time are `x=2t` and `y=4t`.Here `x` and `y` are in metre and `t` in second. Then
The distance travelled by the particle in a time from `t=0` to `t=2s` is ………`m`

Options

A. `2sqrt(3)`
B. `4sqrt(5)`
C. `sqrt(2)`
D. `3sqrt(40)`

Answer

To find the distance traveled by the particle from \( t = 0 \) to \( t = 2 \) seconds, we can follow these steps:

### Step 1: Determine the coordinates at \( t = 0 \) and \( t = 2 \)

The coordinates of the particle are given by:
- \( x(t) = 2t \)
- \( y(t) = 4t \)

**At \( t = 0 \)**:
- \( x(0) = 2 \times 0 = 0 \)
- \( y(0) = 4 \times 0 = 0 \)

So, the coordinates at \( t = 0 \) are \( (0, 0) \).

**At \( t = 2 \)**:
- \( x(2) = 2 \times 2 = 4 \)
- \( y(2) = 4 \times 2 = 8 \)

So, the coordinates at \( t = 2 \) are \( (4, 8) \).

### Step 2: Calculate the distance traveled

The distance traveled by the particle can be calculated using the distance formula between two points in the Cartesian plane:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
- \( (x_1, y_1) = (0, 0) \)
- \( (x_2, y_2) = (4, 8) \)

The distance becomes:
\[ \text{Distance} = \sqrt{(4 - 0)^2 + (8 - 0)^2} \]
\[ = \sqrt{4^2 + 8^2} \]
\[ = \sqrt{16 + 64} \]
\[ = \sqrt{80} \]

### Step 3: Simplify the distance

Now, we simplify \( \sqrt{80} \):
\[ \sqrt{80} = \sqrt{16 \times 5} = \sqrt{16} \times \sqrt{5} = 4\sqrt{5} \]

### Conclusion

Thus, the distance traveled by the particle from \( t = 0 \) to \( t = 2 \) seconds is:
\[ \text{Distance} = 4\sqrt{5} \text{ m} \]

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