A cyclist starts from rest from a point $A$ and travels in a straight line until she comes to a rest at a point $B$. During the motion, her velocity, $v$ m/s, is given by $v = 12t - \frac{3}{2}t^2$, where $t$ is the time in seconds after leaving $A$.

1. Set velocity to zero to find when the cyclist comes to rest

At point $B$, the cyclist comes to rest, so $v = 0$.
$12t - \frac{3}{2}t^2 = 0$

2. Solve the quadratic equation for $t$

$t\left(12 - \frac{3}{2}t\right) = 0$
$t = 0$ (at point $A$) or $12 - \frac{3}{2}t = 0$
$\frac{3}{2}t = 12$
$t = 8$

1. Find the time when the cyclist reaches point $B$ (from part a)

At point $B$, the cyclist comes to a rest, so $v = 0$.
$12t - \frac{3}{2}t^2 = 0$
$t\left(12 - \frac{3}{2}t\right) = 0$
Since $t > 0$ at $B$, $t = 8$.

2. Set up the definite integral to find the distance

Distance $AB = \int_{0}^{8} v \, dt$
$= \int_{0}^{8} \left(12t - \frac{3}{2}t^2\right) dt$

3. Evaluate the definite integral

$AB = \left[ 6t^2 - \frac{1}{2}t^3 \right]_{0}^{8}$
$= \left( 6(8)^2 - \frac{1}{2}(8)^3 \right) - (0)$
$= 6(64) - \frac{1}{2}(512)$
$= 384 - 256 = 128$

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

$a = \frac{dv}{dt}$
$a = \frac{d}{dt}\left(12t - \frac{3}{2}t^2\right)$
$a = 12 - 3t$

2. Substitute $t = 5$ to find the acceleration

When $t = 5$,
$a = 12 - 3(5)$
$a = 12 - 15$
$a = -3$