Question

a boy walks to his school at a distance of 6 km with constant speed of 2.5 km/hr &walks back with constant speed 4km/hr his average speed for round trip expressed in terms of km/hr

Answer 1

Hii..
Here is your answer..

Solution:-
Distance travelled first(D1)=6km
Speed(S1)=2.5km/h

So,
$time = \frac{distance}{speed } \\ = > \frac{6}{2.5} \\ = > 2.4hours$
Now,
Distance travelled second time (D2)=6km
Speed(S2)=4km/h
So,
$time \: = \frac{6}{4} \\ = > 1.5hours$
Therefore,
Total distance=6+6
$= > 12km$
Total time =2.4+1.5
$= > 3.9$
Now,
$average \: speed = \frac{total \: distance}{total \: time} \\ = > \frac{12}{3.9} \\ = > 3.076$
Hence,the average speed=3.076km/h

Hope it helps uh...✌️✌️✌️