A lorry of mass $16000{\text{ }}kg$ climbs a straight hill $ABCD$ which makes an angle $\theta $ with the horizontal, where $\sin \theta  = \frac{1}{{20}}$. For the motion from $A$ to $B$, the work done by the driving force of the lorry is $1200{\text{ }}kJ$ and the resistance to motion is constant and equal to $1240{\text{ }}N$. The speed of the lorry is $15{\text{ }}m{\text{ }}{s^{ - 1}}$ at $A$ and $12{\text{ }}m{\text{ }}{s^{ - 1}}$ at $B$.

a) Find the distance $AB$.

For the motion from $B$ to $D$ the gain in potential energy of the lorry is $2400{\text{ }}kJ$.

b) Find the distance $BD$.

For the motion from $B$ to $D$ the driving force of the lorry is constant and equal to $7200{\text{ }}N$. From $B$ to $C$ the resistance to motion is constant and equal to $1240{\text{ }}N$ and from $C$ to $D$ the resistance to motion is constant and equal to $1860{\text{ }}N$.

c) Given that the speed of the lorry at $D$ is $7{\text{ }}m{\text{ }}{s^{ - 1}}$, find the distance $BC$.