Question

The table below shows data of sprints of animals that traveled 75 meters. At each distance marker, the animals' times were recorded. Which animal is showing the greatest average acceleration from the 25-meter mark to the 50-meter mark?

Answer 1

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

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

$a = 0.076 m/s^2$

Answer 2

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.