Calculate the average velocity of a particle whose position vector changes from $\vec{r}_1 = 5\hat{i}+6\hat{j}$ to $\vec{r}_2 = 2\hat{i}+3\hat{j}$ in a time of 5 seconds.

Calculate the average velocity of a particle whose position vector changes from vecr₁ = 5hati+6hatj to vecr₂ =

IV. Exercises · Q3

\vec{r}_1 = 5\hat{i}+6\hat{j}

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Concept understanding — Average Speed vs Velocity

Average Speed vs Velocity: The Intuition First

Imagine you're walking home from school. You take a shortcut through a park, then stop to buy a snack, then realise you forgot something and run back a bit, then finally walk home. By the time you reach your front door, you've walked a total of 2 km — but your house is only 500 metres from school in a straight line.

That difference — between the total ground you covered and how far you actually ended up from where you started — is the entire point of speed vs velocity.

The Precise Definitions

Average speed is a measure of how fast something is moving overall. It cares only about the total distance travelled, not the direction.

\text{Average speed} = \frac{\text{Total distance travelled}}{\text{Total time taken}}

Average velocity is a measure of how fast and in what direction something is moving overall. It cares about the net displacement — the straight-line distance from start to finish, with a direction.

\text{Average velocity} = \frac{\text{Displacement}}{\text{Total time taken}}

Note

Displacement is the straight-line distance from the starting point to the ending point, with a direction. Distance is the total length of the actual path travelled, with no direction.

The Key Difference in One Sentence

Speed is a scalar (just a number, like 5 km/h). Velocity is a vector (a number and a direction, like 5 km/h north).

That one word — direction — changes everything.

A Concrete Example

You walk 3 km east, then 4 km north. The whole trip takes 1 hour.

Total distance travelled = 3 + 4 = 7 km
Displacement = straight line from start to finish = $\sqrt{3^2 + 4^2} = 5$ km, northeast

Now compute:

\text{Average speed} = \frac{7 \text{ km}}{1 \text{ h}} = 7 \text{ km/h}

\text{Average velocity} = \frac{5 \text{ km, northeast}}{1 \text{ h}} = 5 \text{ km/h, northeast}

Watch out

A common mistake: students think average velocity is just "speed with direction". It's not. It's displacement divided by time, not distance divided by time. If you walk in a circle and return to your starting point, your displacement is zero — so your average velocity is zero, even though your average speed is positive.

When Are They Equal?

Only when the motion is in a straight line without changing direction. If you walk 2 km east in a straight line, then distance = displacement, so average speed = magnitude of average velocity.

But the moment you turn, or stop, or go backwards — they diverge.

Why This Matters for Exams

In Indian board exams (CBSE, ICSE, state boards), you will be asked to:

Distinguish between speed and velocity (scalar vs vector)
Calculate average speed and average velocity from given data
Interpret situations where velocity is zero but speed is not (like a round trip)

Important

Always check: does the problem give you distance or displacement? If it says "returns to starting point", displacement = 0, so average velocity = 0 regardless of how fast the object moved.

The Bottom Line

Quantity	Type	Formula	Depends on
Average speed	Scalar	$\frac{\text{total distance}}{\text{total time}}$	Path taken
Average velocity	Vector	$\frac{\text{displacement}}{\text{total time}}$	Start and end points only

Average speed tells you how fast the journey was. Average velocity tells you how effectively you moved from where you started to where you ended.

Searches for "Average Speed vs Velocity notes class 11" and "Average Speed vs Velocity important questions" both point back to this same core idea, since Average Speed vs Velocity sits squarely within the Motion in a Straight Line coverage of NCERT Class 11 Physics, so it is fair game for both CBSE board questions and competitive-exam numericals. The clearest way to build exam confidence here is to combine this explanation with the NCERT Physics textbook's own solved examples and chapter-end questions.

Step 1. Given $\vec r_1=5\hat i+6\hat j$ , $\vec r_2=2\hat i+3\hat j$ , $\Delta t=5$ s.

Step 2. $\Delta\vec r=\vec r_2-\vec r_1=(2-5)\hat i+(3-6)\hat j=-3\hat i-3\hat j$ .

Step 3. $\vec v_{avg}=\dfrac{\Delta\vec r}{\Delta t}=\dfrac{-3\hat i-3\hat j}{5}=-0.6\hat i-0.6\hat j$ m s $^{-1}$ .

✓Final answer

\vec v_{avg}=-\dfrac35(\hat i+\hat j)=-0.6\hat i-0.6\hat j

←Previous question · 2.11.6Q2 — A particle's position moves from vecr₁ = 3hati+4hatj to vecr₂ = hati+2…→Next question · 2.11.6Q4 — Convert the vector vecr = 3hati+2hatj into a unit vector.

Q.Calculate the average velocity of a particle whose position vector changes from r⃗1=5i^+6j^\vec{r}_1 = 5\hat{i}+6\hat{j}r1​=5i^+6j^​ to r⃗2=2i^+3j^\vec{r}_2 = 2\hat{i}+3\hat{j}r2​=2i^+3j^​ in a time of 5 seconds.

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Q.Calculate the average velocity of a particle whose position vector changes from $\vec{r}_1 = 5\hat{i}+6\hat{j}$ to $\vec{r}_2 = 2\hat{i}+3\hat{j}$ in a time of 5 seconds.

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