A student walks to his school at a distance of 8 km with a constant speed of 2 km/h and walks back at a constant of 4 km/h. Calculate the average speed of the round trip.

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Solution

Step 1:Given data:
Distance of the school, d = 8 km.
On the way, the speed of the boy, $v_{1} = 2 k m / h .$
On the way back, the speed of the boy, $v_{2} = 4 k m / h .$
Let, the average speed of the boy is $v_{a v g}$ .
Step 2: Formula used
$A v e r a g e s p e e d = \frac{T o t a l D i s \tan c e}{T o t a l t i m e}, T i m e = \frac{D i s \tan c e}{S p e e d} .$
Step 3:Finding the value of time:
Now, the total time taken by the boy in going and coming back is
$t = \frac{8}{v_{1}} + \frac{8}{v_{2}} = \frac{8}{2} + \frac{8}{4} = 6 . o r t = 6 h .$
and the total distance traveled by the boy is = 8 + 8 =16 km.
Step 4:Finding the average speed:
$A v e r a g e s p e e d = \frac{16}{6} = 2.66 . o r v_{a v g} = 2.66 k m / h .$
Therefore, the average speed of the boy is 2.66 km/h.

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