Question

Animal "one" accelerates from rest to 18.0 mph in 11.4 s. Animal "two" accelerates from rest to 5.6 mph in the same amount of time.

By what factor is the average acceleration developed by animal "one" greater than the average acceleration developed by animal "two"?

Answer 1

Given initial velocity of animal 1 v₀ = 0m/s

final velocity of animal 1,v_f = 18 mph = 8.04672 m/s

time taken t_f = 11.4 s

average acceleration = v_f -v₀/t_f

av. a = (8.04672-0)/11.4

a₁ = 0.706 m/s²

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initial velocity of animal 2 v₀ = 0m/s

final velocity of animal 2,v_f = 5.6 mph = 2.5034 m/s

time taken t_f = 11.4 s

average acceleration = v_f -v₀/t_f

av. a = (2.5034 -0)/11.4

a₂ = 0.2196 m/s²

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ANSWER: , average acceleration of animal 1 is 3.215 times greater than that of animal 2.

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