name______ date______ per______ for each of the scenarios below, use the initial position and velocity to…

Explanation:

Step1: Recall position - time formula

The formula for position as a function of time in one - dimensional motion with constant velocity is $x = x_i+vt$, 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 1 - second interval.

Step2: Solve for scenario 5

Given $x_i = 0m$ and $v = 5m/s$. At $t = 1s$, $x_1=x_i + vt=0 + 5\times1=5m$ At $t = 2s$, $x_2=0+5\times2 = 10m$

Step3: Solve for scenario 6

Given $x_i = 9m$ and $v=-2m/s$. At $t = 1s$, $x_1=x_i+vt=9+( - 2)\times1=7m$ At $t = 2s$, $x_2=9+( - 2)\times2=9 - 4 = 5m$

Step4: Solve for scenario 7

Given $x_i = 1.5m$ and $v = 2m/s$. At $t = 1s$, $x_1=x_i+vt=1.5+2\times1=3.5m$ At $t = 2s$, $x_2=1.5+2\times2=1.5 + 4 = 5.5m$

Step5: Solve for scenario 8

Given $x_i = 10m$ and $v=-2.5m/s$. At $t = 1s$, $x_1=x_i+vt=10+( - 2.5)\times1=7.5m$ At $t = 2s$, $x_2=10+( - 2.5)\times2=10 - 5 = 5m$

Step6: Solve for scenario 9

Given $x_i = 5m$ and $v=-0.75m/s$. At $t = 1s$, $x_1=x_i+vt=5+( - 0.75)\times1=4.25m$ At $t = 2s$, $x_2=5+( - 0.75)\times2=5 - 1.5 = 3.5m$

The motion maps would have dots placed at the calculated positions at each 1 - second interval. For example, for scenario 5, a dot at $x = 0m$ (initial position), then a dot at $x = 5m$ at $t = 1s$, and a dot at $x = 10m$ at $t = 2s$ and so on for other scenarios.

Answer:

For scenario 5: Positions at $t = 0s,1s,2s$ are $0m,5m,10m$ For scenario 6: Positions at $t = 0s,1s,2s$ are $9m,7m,5m$ For scenario 7: Positions at $t = 0s,1s,2s$ are $1.5m,3.5m,5.5m$ For scenario 8: Positions at $t = 0s,1s,2s$ are $10m,7.5m,5m$ For scenario 9: Positions at $t = 0s,1s,2s$ are $5m,4.25m,3.5m$