A solid horizontal surface is covered with a thin layer of oil. A rectangular block of mass $m = 0.4\,\text{kg}$ is at rest on this surface. An impulse of $1.0\,\text{Ns}$ is applied to the block at time $t = 0$ so that it starts moving along the $x$-axis with a velocity $v(t) = v_{0}e^{-t/\tau}$, where $v_{0}$ is a constant and $\tau = 4\,\text{s}$. The displacement of the block, in metres, at $t = \tau$ is ________. Take $e^{-1} = 0.37$.

Step 1: Identify the given parameters.

The problem provides the following physical values:

$$m = 0.4\,\text{kg}$$ $$J = 1.0\,\text{Ns}$$ $$\tau = 4\,\text{s}$$ $$v(t) = v_{0}e^{-t/\tau}$$ $$e^{-1} = 0.37$$

Step 2: Determine the initial velocity.

The impulse applied at $t = 0$ is equal to the change in momentum of the block.

$$J = m v_{0}$$ $$v_{0} = \frac{J}{m}$$ $$v_{0} = \frac{1.0}{0.4}$$ $$v_{0} = 2.5\,\text{m/s}$$

Step 3: Set up the integral for displacement.

The displacement $x(\tau)$ at time $t = \tau$ is the integral of the velocity function from $t = 0$ to $t = \tau$.

$$x(\tau) = \int_{0}^{\tau} v(t)\,dt$$ $$x(\tau) = \int_{0}^{\tau} v_{0}e^{-t/\tau}\,dt$$

Step 4: Evaluate the integral.

Integrate the exponential function.

$$x(\tau) = \left[ -v_{0}\tau e^{-t/\tau} \right]_{0}^{\tau}$$ $$x(\tau) = v_{0}\tau\left(1 - e^{-1}\right)$$

Step 5: Calculate the final displacement.

Substitute the known values into the integrated expression.

$$x(\tau) = 2.5 \times 4 \times (1 - 0.37)$$ $$x(\tau) = 10 \times 0.63$$ $$x(\tau) = 6.3\,\text{m}$$

Answer: $6.3\,\text{m}$