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nordsb [41]
3 years ago
6

The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recor

ded. Which animal is showing the greatest average acceleration from the 25-meter mark to the 50-meter mark?

Physics
2 answers:
notka56 [123]3 years ago
7 0

Answer:

Animal 2 has maximum acceleration from x = 25 to x = 50

Explanation:

For animal 1

Average velocity at t = 3.5 s

v_1 = \frac{50}{3.5} = 14.3 m/s

Average velocity at t = 3 s

v_1 = \frac{25}{3} = 8.33 m/s

so average acceleration is given as

a = \frac{14.3 - 8.33}{3.5 - 3}

a = 11.9 m/s^2

For animal 2

Average velocity at t = 5.5 s

v_1 = \frac{50}{5.5} = 9.09 m/s

Average velocity at t = 4.5 s

v_1 = \frac{25}{4.5} = 5.55 m/s

so average acceleration is given as

a = \frac{9.09 - 5.55}{5.5 - 4.5}

a = 3.53 m/s^2

For animal 3

Average velocity at t = 9 s

v_1 = \frac{50}{9} = 5.55 m/s

Average velocity at t = 7 s

v_1 = \frac{25}{7} = 3.57 m/s

so average acceleration is given as

a = \frac{5.55 - 3.57}{9 - 7}

a = 0.99 m/s^2

For animal 4

Average velocity at t = 11 s

v_1 = \frac{50}{11} = 4.54 m/s

Average velocity at t = 6 s

v_1 = \frac{25}{6} = 4.16 m/s

so average acceleration is given as

a = \frac{4.54 - 4.16}{11 - 6}

a = 0.076 m/s^2

mr_godi [17]3 years ago
5 0

Answer:

Explanation:

Animal 1 because it takes 3s to go 25 meters 3.5s to go 50 meters and 5s to go 75 meters while the others take longer.

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