Question

a man went from his house to market 3km away with uniform speed 4km/hr and on finding the market closed he insta
ntly turn back with velocity of 6km/h ,find average speed of his journey
​

Answer 1

Given:

A man went from his house to market 3km away with uniform speed 4km/hr and on finding the market closed he instantly turned back with velocity of 6km/h.

To find:

Average speed of his journey

Calculation:

Average speed as the ratio of the total distance travelled by a particle to the time taken to cover that distance.

$\rm{avg. \: v = \dfrac{total \: distance}{total \: time} }$

$= > \rm{avg. \: v = \dfrac{d1 + d2}{t1 +t2} }$

$= > \rm{avg. \: v = \dfrac{3 + 3}{ ( \frac{3}{4}) + (\frac{3}{6}) } }$

$= > \rm{avg. \: v = \dfrac{6}{ ( \frac{3}{4}) + (\frac{3}{6}) } }$

$= > \rm{avg. \: v = \dfrac{2}{ ( \frac{1}{4}) + (\frac{1}{6}) } }$

$= > \rm{avg. \: v = \dfrac{2}{ ( \frac{3 + 2}{12}) } }$

$= > \rm{avg. \: v = \dfrac{2}{ ( \frac{5}{12}) } }$

$= > \rm{avg. \: v = \dfrac{2 \times 12}{ 5} }$

$= > \rm{avg. \: v = \dfrac{24}{ 5} }$

$= > \rm{avg. \: v = 4.8 \: km {hr}^{ - 1} }$

So, final answer is:

$\boxed{\rm{avg. \: v = 4.8 \: km {hr}^{ - 1} }}$