What are vector and scalar quantities? To which class do: momentum, current, electrical energy, acceleration due to gravity and mass belong?

For the automobile (starting from rest):
Initial velocity, $u_A = 0 \text{ ms}^{-1}$
Acceleration, $a_A = 3 \text{ ms}^{-2}$
Distance covered, $s_A = u_A t + \frac{1}{2} a_A t^2 = 0 \cdot t + \frac{1}{2} (3) t^2 = \frac{3}{2} t^2$
Final velocity, $v_A = u_A + a_A t = 0 + 3t = 3t$

For the truck (constant velocity):
Velocity, $v_T = 12 \text{ ms}^{-1}$
Acceleration, $a_T = 0 \text{ ms}^{-2}$
Distance covered, $s_T = v_T t = 12t$

The automobile overtakes the truck when their distances covered are equal:
$s_A = s_T$
$\frac{3}{2} t^2 = 12t$
Since $t \neq 0$ (as they start at the same time and the automobile needs time to catch up), we can divide by $t$:
$\frac{3}{2} t = 12$
$3t = 24$
$t = \frac{24}{3} = 8 \text{ s}$

Using the truck's distance formula (or automobile's):
$s_T = 12t = 12 \times 8 = 96 \text{ m}$
So, the automobile will overtake the truck 96 m beyond its starting point.

Using the automobile's final velocity formula:
$v_A = 3t = 3 \times 8 = 24 \text{ ms}^{-1}$
So, the automobile will be travelling at 24 ms⁻¹ at that time.

The automobile will overtake the truck 96 m beyond its starting point and will be travelling at $\boxed{24 \text{ ms}^{-1}}$ at that time.

11b (i) Define centripetal force and derive its expression.
Centripetal force is the force that acts on an object moving in a circular path and is directed towards the center of the circle. It is responsible for continuously changing the direction of the object's velocity, thus keeping it in circular motion.

Derivation of Centripetal Force Expression:
Consider an object of mass $m$ moving with a constant speed $v$ in a circular path of radius $r$.
The centripetal acceleration $a_c$ is given by:

According to Newton's second law of motion, force $F$ is equal to mass $m$ times acceleration $a$:

Substituting the expression for centripetal acceleration into Newton's second law, we get the expression for centripetal force $F_c$:

11b (ii) A bend in a level road has a radius of $100 \text{ m}$. Find the maximum speed which a car turning this bend may have without skidding if the coefficient of friction between the road and the car is 0.25.

For a car to turn a bend without skidding, the centripetal force required for circular motion must be provided by the static friction force between the tires and the road.
Centripetal force, $F_c = \frac{mv^2}{r}$
Maximum static friction force, $F_f = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

On a level road, the normal force $N$ is equal to the gravitational force $mg$:
$N = mg$
So, $F_f = \mu_s mg$

For no skidding, $F_c \le F_f$. For maximum speed, $F_c = F_f$:
$\frac{mv^2}{r} = \mu_s mg$

Cancel out the mass $m$ from both sides:
$\frac{v^2}{r} = \mu_s g$
$v^2 = \mu_s g r$
$v = \sqrt{\mu_s g r}$

Given:
Radius, $r = 100 \text{ m}$
Coefficient of friction, $\mu_s = 0.25$
Acceleration due to gravity, $g = 9.8 \text{ ms}^{-2}$ (standard value)

$v = \sqrt{0.25 \times 9.8 \text{ ms}^{-2} \times 100 \text{ m}}$
$v = \sqrt{245 \text{ m}^2\text{s}^{-2}}$
$v \approx 15.65 \text{ ms}^{-1}$

The maximum speed the car can have without skidding is $\boxed{15.65 \text{ ms}^{-1}}$.