A passenger arriving in a new town wishes to go from the station to a hotel located $10\ \text{km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\ \text{km}$ long and reaches the hotel in $28\ \text{min}$. What is (a) the average speed of the taxi, (b) the magnitude of average velocity? Are the two equal?

A passenger arriving in a new town wishes to go from the station to a hotel located 10km away on a straight ro

Exercises · 3.11

10\ \text{km}

(a) the average speed of the taxi,

(b) the magnitude of average velocity? Are the two equal?

Sikkim CbseNCERTSubjective· 3mImportance★★★★★est

29% · 20/68 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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 …

Start your 14-day free trial to unlock the full solution →

Average speed depends on the actual path travelled ( $23\,\text{km}$ ), while average velocity depends only on the net displacement ( $10\,\text{km}$ ). They differ: the taxi's average speed is $\approx 49.3\,\text{km/h}$ and the magnitude of average velocity is $\approx 21.4\,\text{km/h}$ .

Why average speed and average velocity differ

Speed is a scalar—it cares only about how much ground you cover, regardless of direction. Velocity is a vector—it cares about where you end up relative to where you started. When a path is not straight, the distance travelled exceeds the straight-line displacement, so average speed will be larger than the magnitude of average velocity.

In this problem, the cabman takes a winding $23\,\text{km}$ route to cover a straight-line distance of only $10\,\text{km}$ . The passenger pays for the longer journey, but physically he's displaced by much less.

Step-by-step solution

1. Convert the time to consistent units

The journey takes $28\,\text{min}$ . Since speeds are typically expressed in $\text{km/h}$ , convert:

t = 28\,\text{min} = \frac{28}{60}\,\text{h} = \frac{7}{15}\,\text{h}

2. Calculate the average speed

Average speed is defined as the total distance travelled divided by the total time:

v_{\text{avg, speed}} = \frac{\text{total distance}}{\text{time}} = \frac{23\,\text{km}}{\frac{7}{15}\,\text{h}} = 23 \times \frac{15}{7} = \frac{345}{7}\,\text{km/h} \approx 49.3\,\text{km/h}

3. Calculate the magnitude of average velocity

Average velocity is defined as the displacement (straight-line distance from start to finish) divided by time. The hotel is $10\,\text{km}$ from the station along a straight road, so the displacement is $10\,\text{km}$ :

|\vec{v}_{\text{avg}}| = \frac{\text{displacement}}{\text{time}} = \frac{10\,\text{km}}{\frac{7}{15}\,\text{h}} = 10 \times \frac{15}{7} = \frac{150}{7}\,\text{km/h} \approx 21.4\,\text{km/h}

4. Compare the two …

Start your 14-day free trial to unlock the full solution →

Concept: Work Entirely in SI Units First, Convert to km/h Only at the End

Method: Metres-and-Seconds Route (No Fractional Hours), Using the $\times 3.6$ Shortcut Last

Both existing solutions convert the $28\ \text{min}$ trip time into a fraction of an hour ( $7/15\ \text{h}$ ) and keep every calculation in kilometres and hours throughout. This method instead converts everything to base SI units (metres, seconds) first, computes both speeds cleanly in m/s, and only converts to km/h as the very last step — avoiding fractional-hour arithmetic entirely.

Step 1 — Convert every given quantity to SI units

\text{distance travelled} = 23\ \text{km} = 23{,}000\ \text{m}, \qquad \text{displacement} = 10\ \text{km} = 10{,}000\ \text{m}

\text{time} = 28\ \text{min} = 28\times60\ \text{s} = 1680\ \text{s}

Step 2 — Average speed, directly in m/s

v_{\text{avg,speed}} = \frac{23{,}000\ \text{m}}{1680\ \text{s}} \approx 13.69\ \text{m/s}

Step 3 — Magnitude of average velocity, directly in m/s

|\vec v_{\text{avg}}| = \frac{10{,}000\ \text{m}}{1680\ \text{s}} \approx 5.95\ \text{m/s}

Step 4 — Convert both to km/h in one shot, at the very end

The conversion factor from m/s to km/h is exactly $3.6$ (i.e. $\times\frac{18}{5}$ ):

13.69\ \text{m/s} \times 3.6 = 49.3\ \text{km/h}, \qquad 5.95\ \text{m/s}\times 3.6 = 21.4\ \text{km/h}

matching $\dfrac{345}{7}\approx49.3\ \text{km/h}$ and $\dfrac{150}{7}\approx21.4\ \text{km/h}$ exactly.

Step 5 — Compare …

Start your 14-day free trial to unlock the full solution →

←Previous question3.10 — On an open ground, a motorist follows a track that turns to his left b…→Next question3.12 — The ceiling of a long hall is 25m high. What is the maximum horizontal…

Q.A passenger arriving in a new town wishes to go from the station to a hotel located 10 km10\ \text{km}10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km23\ \text{km}23 km long and reaches the hotel in 28 min28\ \text{min}28 min. What is

Why average speed and average velocity differ

Step-by-step solution

🎓Unlock everything free for 14 days

Q.A passenger arriving in a new town wishes to go from the station to a hotel located $10\ \text{km}$ away on a straight road from the station. A dishonest cabman takes him along a circuitous path $23\ \text{km}$ long and reaches the hotel in $28\ \text{min}$ . What is

Unlock everything free for 14 days