Sovi.AI - AI Math Tutor

Scan to solve math questions

QUESTION IMAGE

you and a partner are given the following question to test in lab. ques…

Question

you and a partner are given the following question to test in lab. question: if you drop a rock every 10 seconds while running, should there be more, less, or the same amount of distance between the rocks if you are accelerating? your partner makes the following claim: partner’s claim: there should be less distance between the rocks because you are faster. after running the experiment, you and your partner collect the following data: interval | distance rock 1 | 5 meters rock 2 | 10 meters rock 3 | 20 meters based on the data above, write a cer explaining if your partner’s claim is supported or refuted. (claim (1) sentence(s).)

Explanation:

Brief Explanations

The problem involves analyzing the motion of falling rocks, which relates to the principles of physics (specifically kinematics, as it deals with the distance fallen over time intervals). The partner's claim about the distance between rocks when accelerating (due to gravity) can be evaluated using the equations of motion for free fall. In free fall, the distance fallen $d = \frac{1}{2}gt^2$, where $g$ is acceleration due to gravity and $t$ is time. As time increases, the distance fallen in each subsequent interval increases, so the distance between rocks (which fall at different start times) should increase, meaning the partner's claim (less distance between rocks) is incorrect. To support or refute, we use the free - fall kinematic equation and the relationship between time intervals and distance fallen.

Answer:

The partner's claim is incorrect. When an object is in free fall (accelerating due to gravity), the distance it falls is given by the equation $d=\frac{1}{2}gt^{2}$. For a rock dropped at time $t = 0$, at $t = 5$ s, $d_1=\frac{1}{2}g(5)^{2}$; at $t = 10$ s, $d_2=\frac{1}{2}g(10)^{2}$; at $t = 20$ s, $d_3=\frac{1}{2}g(20)^{2}$. The distance between Rock 1 and Rock 2 at $t = 10$ s is $d_2 - d_1=\frac{1}{2}g(10^{2}-5^{2})=\frac{1}{2}g(75)$, and between Rock 2 and Rock 3 at $t = 20$ s is $d_3 - d_2=\frac{1}{2}g(20^{2}-10^{2})=\frac{1}{2}g(300)$. Since $300>75$, the distance between the rocks increases as time (and thus the interval) increases. So the distance between the rocks should be more, not less, when accelerating.