LP Grade 10 Physical Sciences...

This shows that $\boxed{\text{72 km} \cdot \text{h}^{-1} = 20 \text{ m} \cdot \text{s}^{-1}}$.

Initial speed ($v$) = $20 \text{ m} \cdot \text{s}^{-1}$ (from 4.2)
Reaction time ($\Delta t$) = $0.4 \text{ s}$

distance = speed $\times$ time.

The distance travelled during the reaction time is $\boxed{\text{8 m}}$.

Initial velocity ($u$) = $20 \text{ m} \cdot \text{s}^{-1}$
Final velocity ($v$) = $0 \text{ m} \cdot \text{s}^{-1}$ (car comes to rest)
Acceleration ($a$) = $-2.5 \text{ m} \cdot \text{s}^{-2}$ (deceleration)
Using the equation of motion: $v = u + a \Delta t_{\text{braking}}$

The total time taken for the motor car to come to rest is $\boxed{\text{8.4 s}}$.

Initial velocity ($u$) = $20 \text{ m} \cdot \text{s}^{-1}$
Final velocity ($v$) = $0 \text{ m} \cdot \text{s}^{-1}$
Acceleration ($a$) = $-2.5 \text{ m} \cdot \text{s}^{-2}$
Using the equation of motion: $v^2 = u^2 + 2a \Delta x_{\text{braking}}$

The initial distance between the car and the truck is $100 \text{ m}$.
Since $88 \text{ m} < 100 \text{ m}$, the motor car will stop before reaching the wreck of the truck.
The motor car will stop $\boxed{\text{before reaching the wreck of the truck}}$.