A cyclist moves along a straight road from home $O$. His velocity, $v\text{ m/min}$, is given by $v = 25 + 10t - 3t^2$, where $t$ is the time in minutes after leaving $O$. Calculate

1. Find the time when the cyclist is instantaneously at rest by setting $v = 0$

When the cyclist is instantaneously at rest, $v = 0$.
$25 + 10t - 3t^2 = 0$
$3t^2 - 10t - 25 = 0$
$(3t + 5)(t - 5) = 0$
Since $t \ge 0$, $t = 5$.

2. Find the expression for acceleration by differentiating velocity

$a = \frac{dv}{dt}$
$a = \frac{d}{dt}(25 + 10t - 3t^2)$
$a = 10 - 6t$

3. Calculate the acceleration at $t = 5$

When $t = 5$,
$a = 10 - 6(5)$
$a = 10 - 30 = -20$

1. Find the time when the velocity is at a maximum

For maximum velocity, $\frac{dv}{dt} = 0$.
$\frac{dv}{dt} = 10 - 6t$
$10 - 6t = 0$
$t = \frac{10}{6} = \frac{5}{3}$

2. Find the expression for displacement, $s$, by integrating velocity

$s = \int v \, dt$
$s = \int (25 + 10t - 3t^2) \, dt$
$s = 25t + 5t^2 - t^3 + c$
Since the cyclist starts from $O$, $s = 0$ when $t = 0$, so $c = 0$.
$s = 25t + 5t^2 - t^3$

3. Calculate the displacement at the time of maximum velocity

Substitute $t = \frac{5}{3}$ into the displacement equation:
$s = 25\left(\frac{5}{3}\right) + 5\left(\frac{5}{3}\right)^2 - \left(\frac{5}{3}\right)^3$
$s = \frac{125}{3} + 5\left(\frac{25}{9}\right) - \frac{125}{27}$
$s = \frac{1125}{27} + \frac{375}{27} - \frac{125}{27}$
$s = \frac{1375}{27} = 50\frac{25}{27}$

1. Find the time when the cyclist is instantaneously at rest

Set $v = 0$:
$25 + 10t - 3t^2 = 0$
$3t^2 - 10t - 25 = 0$
$(3t + 5)(t - 5) = 0$
$t = 5$ or $t = -\frac{5}{3}$ (reject since $t \ge 0$)

2. Find the displacement function $s(t)$

$s = \int v \, dt = \int (25 + 10t - 3t^2) \, dt$
$s = 25t + 5t^2 - t^3 + c$
Since $s = 0$ when $t = 0$, $c = 0$.
$s = 25t + 5t^2 - t^3$

3. Calculate the total distance travelled in the first 8 minutes

The cyclist changes direction at $t = 5$.
Displacement at $t = 0$: $s(0) = 0$
Displacement at $t = 5$: $s(5) = 25(5) + 5(5)^2 - (5)^3 = 125 + 125 - 125 = 125\text{ m}$
Displacement at $t = 8$: $s(8) = 25(8) + 5(8)^2 - (8)^3 = 200 + 320 - 512 = 8\text{ m}$
Distance travelled from $t=0$ to $t=5$: $|125 - 0| = 125\text{ m}$
Distance travelled from $t=5$ to $t=8$: $|8 - 125| = 117\text{ m}$
Total distance = $125 + 117 = 242\text{ m}$

4. Calculate the average speed

Average speed = $\frac{\text{Total distance}}{\text{Total time}}$
Average speed = $\frac{242}{8} = 30.25\text{ m/min}$