Answer: \(\dfrac{11}{4e}-\dfrac{2}{e^{3/2}}\), approximately \(0.565\text{ m}\).

The particle is never at rest, and the given velocities are positive on the interval considered, so the distance travelled is the integral of the velocity.

From \(t=1\) to \(t=2\),

\(v=\frac{t}{2e}.\)

So

From \(t=2\) to \(t=3\),

\(v=e^{-t/2}.\)

Thus

This gives

\(-2e^{-3/2}+2e^{-1}.\)

Therefore the total distance is

\(\frac3{4e}+2e^{-1}-2e^{-3/2}.\)

Combine the first two terms:

\(\frac3{4e}+\frac2e=\frac{11}{4e}.\)

So the exact distance is

\(\frac{11}{4e}-\frac2{e^{3/2}}.\)

Numerically, this is approximately

\(0.565\text{ m}.\)