Question

name noah lozano date 9/22/25 per 2 unit 2: motion maps for each of the scenarios below, use the initial position and velocity to create a motion map. use a time interval of 1 second. ($\delta t = 1\\ s$) dont forget to label the position axis! 1. $x_i = 0\\ m, v = 3\\ m/s$ 2. $x_i = 0\\ m, v=-3\\ m/s$ 3. $x_i=-2\\ m, v = 4\\ m/s$ 4. $x_i = 12\\ m, v=-2\\ m/s$

Explanation:

Step1: Recall position - velocity relation

The position $x$ at time $t$ is given by the formula $x = x_i+vt$, where $x_i$ is the initial position, $v$ is the velocity and $t$ is the time. Since $\Delta t = 1s$, we can find the position at each second - interval.

Step2: For $x_i = 0m, v = 3m/s$

At $t = 1s$, $x=x_i + vt=0 + 3\times1=3m$. At $t = 2s$, $x=0+3\times2 = 6m$. At $t = 3s$, $x=0 + 3\times3=9m$. And so on. The motion map has an arrow starting at $x = 0$ and pointing to the right (positive direction) with the tip at $x = 3$ for the first - time interval, then from $x = 3$ to $x = 6$ for the second time interval and so on.

Step3: For $x_i = 0m, v=-3m/s$

At $t = 1s$, $x=x_i+vt=0+( - 3)\times1=-3m$. At $t = 2s$, $x=0+( - 3)\times2=-6m$. At $t = 3s$, $x=0+( - 3)\times3=-9m$. The motion map has an arrow starting at $x = 0$ and pointing to the left (negative direction) with the tip at $x=-3$ for the first - time interval, then from $x=-3$ to $x=-6$ for the second time interval and so on.

Step4: For $x_i=-2m, v = 4m/s$

At $t = 1s$, $x=x_i+vt=-2 + 4\times1=2m$. At $t = 2s$, $x=-2+4\times2 = 6m$. At $t = 3s$, $x=-2+4\times3 = 10m$. The motion map has an arrow starting at $x=-2$ and pointing to the right (positive direction) with the tip at $x = 2$ for the first - time interval, then from $x = 2$ to $x = 6$ for the second time interval and so on.

Step5: For $x_i = 12m, v=-2m/s$

At $t = 1s$, $x=x_i+vt=12+( - 2)\times1 = 10m$. At $t = 2s$, $x=12+( - 2)\times2=8m$. At $t = 3s$, $x=12+( - 2)\times3 = 6m$. The motion map has an arrow starting at $x = 12$ and pointing to the left (negative direction) with the tip at $x = 10$ for the first - time interval, then from $x = 10$ to $x = 8$ for the second time interval and so on.

The motion map for $x_i = 12m, v=-2m/s$ should have an arrow starting at $x = 12$ and pointing to the left. The tip of the first arrow should be at $x = 10$, the tip of the second arrow starting from $x = 10$ should be at $x = 8$, the tip of the third arrow starting from $x = 8$ should be at $x = 6$ and so on.

Answer:

The motion map for $x_i = 12m, v=-2m/s$ has arrows pointing to the left (negative direction) starting at $x = 12$, with the positions at $t = 1s,x = 10m$; $t = 2s,x = 8m$; $t = 3s,x = 6m$ etc.