Animal "one" accelerates from rest to 17.3 mph in 19.5 s. Animal "two" accelerates from rest...

Question

Question

Animal "one" accelerates from rest to 17.3 mph in 19.5 s. Animal "two" accelerates from rest to 8.6 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"?

science physics

Answer 1

Answer #1

accleration of the first animal is

a1 = v/t = ( 17.3 mph) ( 0.44704 m/s)/19.5 s =0.396 m/s^2

for second animal

a = v/t = ( 8.6 mph) ( 0.44704 m/s)/19.5 s =0.197 m/s^2

ration

a1/a2 = 0.396 m/s^2/0.197 m/s^2 =2

a1 = 2a2

Answer 2

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