A particle moves towards east from a point $$A$$ to a point $$B$$ at the rate of $$4$$ km/h and then towards north from $$B$$ to $$C$$ at the rate of $$5$$ km/h. If $$AB = 12$$ km and $$BC = 5$$ km, then its average speed for its journey from $$A$$ to $$C$$ and resultant average velocity direct from $$A$$ to $$C$$ are respectively

Let the motion of the particle be broken down into two perpendicular segments.

First segment from point A to point B:

The direction is towards East.

The distance is given by:

$$ AB = 12 \text{ km} $$

The speed for this segment is:

$$ v_1 = 4 \text{ km/h} $$

The time taken for this segment is calculated by dividing distance by speed:

$$ t_1 = \frac{AB}{v_1} = \frac{12}{4} = 3 \text{ hours} $$

Second segment from point B to point C:

The direction is towards North.

The distance is given by:

$$ BC = 5 \text{ km} $$

The speed for this segment is:

$$ v_2 = 5 \text{ km/h} $$

The time taken for this segment is:

$$ t_2 = \frac{BC}{v_2} = \frac{5}{5} = 1 \text{ hour} $$

Now find the total values for the entire journey.

The total distance traveled by the particle is:

$$ \text{Total Distance} = AB + BC = 12 + 5 = 17 \text{ km} $$

The total time taken for the journey is:

$$ \text{Total Time} = t_1 + t_2 = 3 + 1 = 4 \text{ hours} $$

Calculate the average speed of the particle as a fraction in its simplest coprime form:

$$ \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{17}{4} \text{ km/h} $$

To find the average velocity, we need the net displacement from the starting point A to the final point C. Since the movements towards East and North are perpendicular, we use the Pythagorean theorem:

$$ AC = \sqrt{AB^2 + BC^2} $$

$$ AC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ km} $$

Calculate the magnitude of the average velocity as a fraction in its simplest coprime form:

$$ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Total Time}} = \frac{AC}{\text{Total Time}} = \frac{13}{4} \text{ km/h} $$

The direction of this velocity is directed along the straight line from point A to point C.

Final Answer:

The average speed and the resultant average velocity are $$ \frac{17}{4} $$ km/h and $$ \frac{13}{4} $$ km/h respectively.