A paticle moves towards east from a point A to a point B at the rate of 4 km/h and then towards nort

Q: Wait, why can't I just take the average of the two speeds, $\frac{4+5}{2} = 4.5 \text{ km/h}$?

That is a classic trap! Average speed is not the arithmetic mean of the speeds. It is defined as the total distance divided by the total time. Because the particle spends different amounts of time at different speeds, a simple average of the speeds will give you the wrong result. Always stick to the definition: $\text{Total Distance} / \text{Total Time}$.

Q: Why does the displacement use the Pythagorean theorem? Is the path always a triangle?

In this specific problem, the particle moves East and then North, which creates a $90^{\circ}$ angle between the two segments. This forms a right-angled triangle where the displacement is the hypotenuse. If the particle had turned at a different angle, we would use the Law of Cosines, but for a $90^{\circ}$ turn, Pythagoras is your best friend.

Q: Does the direction matter for average speed?

No, average speed is a scalar quantity. It only cares about how much ground you covered, not where you went. Average velocity, however, is a vector quantity; it cares deeply about your starting point and your final destination, regardless of the path you took to get there.

Question

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/hr. 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

Options:

(A) $\frac{13}{9}$ km/h and $\frac{17}{9}$ km/h

(B) $\frac{13}{4}$ km/h and $\frac{17}{4}$ km/h

(C) $\frac{17}{9}$ km/h and $\frac{13}{9}$ km/h

(D) $\frac{17}{4}$ km/h and $\frac{13}{4}$ km/h

Sigma Percentile · Accepted Answer

### The Kinematics of a Journey: Scalar vs. Vector

Welcome, future engineer! Today, we are going to dissect a classic kinematics problem that serves as a gateway to understanding the profound difference between scalar and vector quantities.

Imagine you are standing at point $A$, watching a particle embark on a journey. It moves East to point $B$, and then makes a sharp turn to head North to point $C$. This isn't just a walk; it is a lesson in how we measure motion.

### Phase 1: The Scalar World (Average Speed)

First, let's look at the journey through the lens of a scalar. Scalars only care about magnitude. When we talk about average speed, we are asking: "How much ground did the particle cover in total, and how long did it take?"

We break the journey into two legs. For the first leg, from $A$ to $B$, the particle covers a distance of $12 	ext{ km}$ at a speed of $4 	ext{ km/h}$.

Using the simple relation $	ext{Time} = \frac{	ext{Distance}}{	ext{Speed}}$, we find the time taken for the first leg:
$$t_1 = \frac{12}{4} = 3 	ext{ hours}$$

Then, the particle turns North to reach point $C$. It covers $5 	ext{ km}$ at a speed of $5 	ext{ km/h}$.

Again, applying our formula, the time for this second leg is:
$$t_2 = \frac{5}{5} = 1 	ext{ hour}$$

To find the average speed, we need the total distance and the total time. The total distance is simply the sum of the two legs: $12 	ext{ km} + 5 	ext{ km} = 17 	ext{ km}$.

The total time is $3 	ext{ hours} + 1 	ext{ hour} = 4 	ext{ hours}$. Thus, the average speed is:
$$	ext{Average Speed} = \frac{17}{4} = 4.25 	ext{ km/h}$$

### Phase 2: The Vector World (Average Velocity)

Now, let's shift our perspective. Average velocity is a vector quantity. It doesn't care about the path taken; it only cares about the displacement—the shortest, straight-line distance from the starting point $A$ to the final point $C$.

Because the particle moved East and then turned North, it traced out two sides of a right-angled triangle. The displacement is the hypotenuse of this triangle.

We invoke the Pythagorean theorem:
$$AC = \sqrt{AB^2 + BC^2}$$

Substituting our values, we get:
$$AC = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 	ext{ km}$$

Now, we calculate the average velocity using the formula $	ext{Average Velocity} = \frac{	ext{Displacement}}{	ext{Total Time}}$. We already know the total time is $4 	ext{ hours}$.

Therefore, the average velocity is:
$$	ext{Average Velocity} = \frac{13}{4} = 3.25 	ext{ km/h}$$

### The Synthesis

Look at the beauty of this comparison. The average speed ($4.25 	ext{ km/h}$) tells us about the effort of the journey, while the average velocity ($3.25 	ext{ km/h}$) tells us about the net progress made.

In physics, these two numbers tell very different stories about the same event. By mastering this distinction, you are not just solving a problem; you are learning to see the world through the eyes of a physicist.

**Keep practicing, keep questioning, and keep falling in love with the elegance of these laws!**

Select Answer:

Visualized Solution

Mapping the Journey: Point $A$ to $B$

Calculating Time for $A B$

The Turn: Point $B$ to $C$

Calculating Time for $B C$

Total Journey Parameters

Calculating Average Speed

Finding Displacement $A C$

Atomic Compute: Displacement Value

Calculating Average Velocity

Final Conclusion

The Sigma Insight: Scalar and Vector Quantities

The Kinematics of a Journey

Phase 1

Phase 2

The Synthesis

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