A particle $P$ moves in a straight line, so that, $t$ seconds after leaving a fixed point $O$, its velocity, $v$ m/s, is given by $v = 10e^{-2t} - 3$.

1. Substitute $t = 0$ into the velocity equation to find the initial velocity

When $t = 0$,
$v = 10e^{-2(0)} - 3$
$v = 10(1) - 3$
$v = 7$

1. Find the expression for acceleration by differentiating velocity with respect to time

$a = \frac{dv}{dt} = \frac{d}{dt}(10e^{-2t} - 3) = -20e^{-2t}$

2. Substitute $t = 1$ to find the acceleration

When $t = 1$,
$a = -20e^{-2(1)} = -20e^{-2} \approx -2.71 \text{ m/s}^2$

1. Set velocity to zero for instantaneous rest

For instantaneous rest, $v = 0$.
$10e^{-2t} - 3 = 0$

2. Isolate the exponential term

$10e^{-2t} = 3$
$e^{-2t} = 0.3$

3. Solve for $t$ using natural logarithms

$-2t = \ln 0.3$
$t = -\frac{1}{2} \ln 0.3$
$t \approx 0.601986$
$t \approx 0.602 \text{ (to 3 s.f.)}$

1. Find the time $t$ when the particle is at instantaneous rest (from part iii).

At instantaneous rest, $v = 0$.
$10e^{-2t} - 3 = 0$
$e^{-2t} = 0.3$
$-2t = \ln 0.3$
$t = -\frac{1}{2}\ln 0.3 \approx 0.601986$

2. Set up the definite integral for the distance travelled.

Since $v > 0$ for $0 \leqslant t < 0.601986$, the distance travelled is the integral of $v$ from $t = 0$ to $t = 0.601986$.
$\text{Distance} = \int_0^{0.601986} (10e^{-2t} - 3) \, dt$

3. Integrate the velocity function.

$\int (10e^{-2t} - 3) \, dt = \left[ \frac{10e^{-2t}}{-2} - 3t \right]_0^{0.601986}$
$= \left[ -5e^{-2t} - 3t \right]_0^{0.601986}$

4. Evaluate the definite integral to find the final distance.

$= \left( -5e^{-2(0.601986)} - 3(0.601986) \right) - \left( -5e^0 - 0 \right)$
$= \left( -5(0.3) - 1.805958 \right) - (-5)$
$= -1.5 - 1.805958 + 5$
$= 1.69404 \approx 1.69 \text{ m (to 3 s.f.)}$

1. State the property of the exponential function

For all real values of $t$, $e^{-2t} > 0$.

2. Deduce the inequality for $v$

Since $e^{-2t} > 0$, it follows that $10e^{-2t} > 0$.
Therefore, $v = 10e^{-2t} - 3 > 0 - 3 = -3$.