Two boys come home from school, they walk the same distance of 300 m. They start from school at the same time but one walks home with 1 m / s and the other one on bicycle with a velocity of 16 km / h. How fast gets home the second boy compared to first one?

Two boys come home from school, they walk the same distance of 300 m. They start from school at the same time but one walks home with 1 m / s and the other one on bicycle with a velocity of 16 km / h. How fast gets home the second boy compared to first one?



Solution

boy 1

According to the velocity equation:

v = Δd / Δt

1 m / s = 300 m / Δt

Δt = 300 m / 1m / s

Δt = 300 s

Δt boy 1 = 5 minutes


boy 2

v = Δd / Δt


16 km / h = 300 m / Δt

Δt = 300 m / 16 km / h

Δt = 300 m / 16 * (1000 m / 3,600 s)

Δt = 300 m / 16,000 m / 3,600 s)

Δt = 300 m / 4.4 m / s

Δt = 66.7 s

Δt boy 2 = 1 minute and 7 seconds



Δt boy 1 - Δt boy 2 = 5 minutes - 1 minute and 7 seconds =>

300 s - 66.7 s = 233.3 seconds


The second boy reaches home with 233.3 seconds (3.88 minutes) earlier than the first boy.


Mechanics

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