Question

man travel with a speed of 6 kilometre per hour for first half of the distance and with a speed of 4 kilometre for the next half calculate the average speed​

Answer 1

Let the half distance covered be d and another half distance covered be d.

We know that,

$\bigstar \: \boxed{\sf {Speed_{avg} = \dfrac{Distance_{(Total)}}{Time_{(Total)}} }}$

★ Calculating total distance :

→ Total distance = (d + d) km

→ Total distance = 2d km

★ Calculating total time :

In first half :

• Speed = 6 km/h
• Distance = d km

So,

$\longmapsto \sf { Time \: (t_1) = \dfrac{Distance}{Speed} }$

$\longmapsto \sf { Time \: (t_1) = \dfrac{d}{6} }$ hr

In the another half :

• Speed = 4 km/h
• Distance = d

So,

$\longmapsto \sf { Time \: (t_2) = \dfrac{Distance}{Speed} }$

$\longmapsto \sf { Time \: (t_2) = \dfrac{d}{4} }$ hr

Therefore,

$\longmapsto \sf {Total \: Time= t_1 + t_2}$

$\longmapsto \sf {Total \: Time= \dfrac{d}{6} + \dfrac{d}{4} }$ hr

$\longmapsto \sf {Total \: Time= \dfrac{3d + 2d}{12} }$ hr

$\longmapsto \sf \red{Total \: Time= \dfrac{5d}{12} }$ hr

Now, as we know that :-

$\bigstar \: \boxed{\sf {Speed_{avg} = \dfrac{Distance_{(Total)}}{Time_{(Total)}} }}$

$\longmapsto \sf{Speed_{avg} = 2d \div \dfrac{5d}{12} }$

$\longmapsto \sf{Speed_{avg} = 2\not{d} \times \dfrac{12}{5\not{d}} }$

$\longmapsto \sf{Speed_{avg} = 2 \times \dfrac{12}{5} }$

$\longmapsto \sf{Speed_{avg} = \dfrac{24}{5} }$

$\longmapsto\boxed { \sf \red{Speed_{avg} = 4.8 \: km/h }}$

Therefore, average speed is 4.8 km/h.