$(i)$ Explain clearly, with examples, the distinction between :
(a) magnitude of displacement over an interval of time and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed over the same interval.
Show in both (a) and (b) that the second quantity is either greater than or equal to the first. When is the equality sign true ?

$(i i)$ A man walks on a straight road from his home to a market $2.5 k m$ away with a speed of $5 k m h^{- 1}$ . Finding the market closed, he instantly turns and walks back home with a speed of $7.5 k m h^{- 1}$ . What is the:
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) $0$ to $30 m i n$ , (ii) $0$ to $50 m i n$ , (iii) $0$ to $40 m i n$ ?

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Solution

$i$
a)-The magnitude of displacement over an interval of time is the shortest distance (which is a straight line) between the initial and final positions of the particle
The total path length of a particle is the actual path length covered by the particle in a given interval of time.
For example, suppose a particle moves from point A to point B and then, comes back to a point, C (between A and B) taking a total time $t$ , . Then, the magnitude of displacement of the particle = AC.
Whereas, total path length = AB + BC
b)-Magnitude of velocity = $\frac{M a g n i t u d e o f d i s p l a c e m e n t}{t i m e}$
For given figure average velocity $v = \frac{A C}{t}$
Average speed= $\frac{t o t a l p a t h}{t}$
$S = \frac{A B + B C}{t}$
Since (AB + BC) > AC, average speed is greater than the magnitude of average velocity. The two quantities will be equal if the particle continues to move along a straight line.
$(i i)$
a) Time taken to reach market $t_{1} = \frac{2.5}{5} = 0.5 h o u r = 30 m i n$
time taken to get back to home is $t2=dfrac2.57.5=.33hour=20min$
Average velocity for 0-30 in is $v = \frac{2.5}{.5} = 5 k m / h$
Average speed for 0-30 in is $s = \frac{2.5}{.5} = 5 k m / h$
total time he took for travelling $t = 30 + 20 = 50 m i n = \frac{5}{6} h o u r$
When he reached back then net displacement is zero
so for 0-50 min
Average velocity for 0-50 in is $v = \frac{0}{\frac{5}{6}} = 0 k m / h$
Total distance he traveled when he arrive back is $2.5 + 2.5 = 5 k m$
Average speed for 0-50 in is $v = \frac{5}{\frac{5}{6}} = 6 k m / h$

Distance traveled in first 30 min is =2.5 km
distance traveled (while returning ) in 10 min is $d = 7.5 \times \frac{1}{6} = 1.25 k m$
Total time is 40 min or $\frac{2}{3}$
So displacement in 0-40 min is $2.5 - 1.25 = 1.25 k m$
Average velocity for 0-40 in is $v = \frac{1.25}{\frac{2}{3}} = 1.875 k m / h$
Average speed for 0-40 in is $v = \frac{3.75}{\frac{2}{3}} = 5.625 k m / h$

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Q. A man walks on a straight road from his home to a market

2.5 k m

away with a speed of

5 k m h^{- 1}

. Finding the market closed, he instantly turns and walks back home with a speed of

7.5 k m h^{- 1}

. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time
(i)

0

30 m i n

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(ii)

0

50 m i n

(iii)

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Q. A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h¯¹. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h¯¹. What is the (a) magnitude of average velocity, and (b) average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min ? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero !]

A man walks on a straight road from his home to a market 2.5km away with a speed of 5km/h. Finding the market closed, he instantly turns and walk back home with a speed of 7.5 km/h. The average speed of the man over the, the interval of time 0 to 40min is equal to?

Explain clearly, with examples, the distinction between:
a) magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
b) magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].

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