Question

question 4 (3 points) (02.02 mc) 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? time at distance markers animal 25 meters 50 meters 75 meters animal 1 13 seconds 3.5 seconds 3.5 seconds animal 2 24.5 seconds 5.5 seconds 5.75 seconds animal 3 37 seconds 9 seconds 11 seconds animal 4 46 seconds 11 seconds 15 seconds a animal 1 b animal 2 c animal 3 d animal 4

Explanation:

Step1: Recall acceleration formula

Acceleration $a=\frac{\Delta v}{\Delta t}$. For constant - acceleration motion over a distance, we can also use the concept of average speed change over time. First, find the time taken to move from 25 - meter to 50 - meter mark for each animal.
For Animal 1: Time taken $t_1 = 8.5 - 3.5=5$ seconds.
For Animal 2: Time taken $t_2 = 5.5 - 2.5 = 3$ seconds.
For Animal 3: Time taken $t_3=9 - 4=5$ seconds.
For Animal 4: Time taken $t_4 = 11 - 6=5$ seconds.
The distance from 25 - meter to 50 - meter mark is $\Delta d=50 - 25 = 25$ meters.
We can assume initial velocity $u$ at 25 - meter mark and final velocity $v$ at 50 - meter mark. Using the formula $d=ut+\frac{1}{2}at^{2}$, or in terms of average speed $\bar{v}=\frac{d}{t}$. Since we want to compare accelerations, we can also use the fact that for a fixed distance $\Delta d = 25$m, the animal that takes the least time has the greatest average acceleration (from $d=\bar{v}t$ and $\bar{v}=\frac{u + v}{2}$, and $a=\frac{v - u}{t}$, for a fixed $d$, smaller $t$ implies larger $a$).

Step2: Compare times

We have $t_1 = 5$ seconds, $t_2 = 3$ seconds, $t_3 = 5$ seconds, $t_4 = 5$ seconds.
Since Animal 2 takes the least time to move from the 25 - meter mark to the 50 - meter mark, it has the greatest average acceleration in this interval.

Answer:

b. Animal 2