A particle starts from rest at a fixed point $O$ and moves in a straight line towards a point $A$. Its velocity, $v$ m/s, $t$ seconds after leaving $O$, is given by $v = 6 - 4e^{-2t}$.
The particle reaches $A$ when $t = \ln 2$,

1. Substitute $t = 0$ into $v$ and evaluate

At $O$, $t = 0$:
$v = 6 - 4e^{-2(0)} = 6 - 4e^{0} = 6 - 4(1) = 2$
So the velocity at $O$ is $2$ m/s.

1. Differentiate the velocity function to find the acceleration function

$a = \frac{dv}{dt} = \frac{d}{dt}(6 - 4e^{-2t})$
$a = -4(-2)e^{-2t} = 8e^{-2t}$

2. Substitute $t = \ln 2$ into the acceleration function

At $A$, $t = \ln 2$.
$a = 8e^{-2(\ln 2)}$
$a = 8e^{\ln(2^{-2})}$
$a = 8(2^{-2}) = 8 \times \frac{1}{4} = 2$

1. Analyse the range of the exponential term to determine the minimum possible velocity.

For all $t \geqslant 0$, $e^{-2t} \leqslant 1$ and $e^{-2t} > 0$.
Therefore, $4e^{-2t} \leqslant 4$.
$v = 6 - 4e^{-2t} \geqslant 6 - 4 = 2$.

2. Conclude that the velocity is always positive.

Since $v \geqslant 2$ for all $t \geqslant 0$, $v$ is never equal to $0$. Hence, the particle never comes to rest.

1. Set up the definite integral for the distance from $O$ to $A$.

The distance $s$ is given by the integral of velocity from $t = 0$ to $t = \ln 2$:
$s = \int_0^{\ln 2} (6 - 4e^{-2t}) \, dt$

2. Integrate the velocity function.

$s = \left[ 6t - \frac{4e^{-2t}}{-2} \right]_0^{\ln 2}$
$s = \left[ 6t + 2e^{-2t} \right]_0^{\ln 2}$

3. Evaluate the definite integral using the limits.

Substitute $t = \ln 2$:
$6(\ln 2) + 2e^{-2\ln 2} = 6\ln 2 + 2e^{\ln(2^{-2})} = 6\ln 2 + 2\left(\frac{1}{4}\right) = 6\ln 2 + 0.5$

Substitute $t = 0$:
$6(0) + 2e^0 = 2$

Calculate the difference:
$s = (6\ln 2 + 0.5) - 2$
$s = 6\ln 2 - 1.5$
$s \approx 2.66 \text{ m (to 3 s.f.)}$