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 $40 m i n$ ?

Question

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Answer 1

Distance to market $s = 2.5 k m = 2.5 \times 1 0^{3} = 2500 m$
Speed with which he goes to market $= 5 k m / h = 5 \frac{1 0 ^{3}}{3 6 0 0} = \frac{2 5}{1 8} m / s$
Speed with which he comes back $= 7.5 k m / h = 7.5 \times \frac{1 0 ^{3}}{3 6 0 0} = \frac{7 5}{3 6} m / s$
(a)Average velocity is zero since his displacement is zero.
(b)
(i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market: $\frac{2 . 5}{5} = 1 / 2 h = 30$ minutes.
Average speed over this interval $= 5 k m / h$
(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes : $7.5 \times \frac{1}{3} = 2.5 k m$
His average speed in 0 to 50 minutes: $V_{a v g} = \frac{d i s t a n c e t r a v e l e d}{t i m e}$
$= \frac{2 . 5 + 2 . 5}{( 5 0 / 6 0 )} = 6 k m / h$
(iii)In 40-30=10 minutes he travels a distance of : $7.5 \times \frac{1}{6} = 1.25 k m$
$V_{a v g} = \frac{2 . 5 + 1 . 2 5}{( 4 0 / 6 0 )} = 5.625 k m / h$

Answer 2

Distance to market

s = 2.5 k m = 2.5 \times 1 0^{3} = 2500 m

Speed with which he goes to market

= 5 k m / h = 5 \frac{1 0 ^{3}}{3 6 0 0} = \frac{2 5}{1 8} m / s

Speed with which he comes back

= 7.5 k m / h = 7.5 \times \frac{1 0 ^{3}}{3 6 0 0} = \frac{7 5}{3 6} m / s

(a)Average velocity is zero since his displacement is zero.

(b)

(i)Since the initial speed is 5km/s and the market is 2.5 km away,time taken to reach market:

\frac{2 . 5}{5} = 1 / 2 h = 30

minutes.

Average speed over this interval

= 5 k m / h

(ii)After 30 minutes,the man is travelling wuth 7.5 km/h speed for 50-30=20 minutes.The distance he covers in 20 minutes :

7.5 \times \frac{1}{3} = 2.5 k m

His average speed in 0 to 50 minutes:

V_{a v g} = \frac{d i s t a n c e t r a v e l e d}{t i m e}

= \frac{2 . 5 + 2 . 5}{( 5 0 / 6 0 )} = 6 k m / h

(iii)In 40-30=10 minutes he travels a distance of :

7.5 \times \frac{1}{6} = 1.25 k m

V_{a v g} = \frac{2 . 5 + 1 . 2 5}{( 4 0 / 6 0 )} = 5.625 k m / h