Question

1. the graph below is a graph of position versus time. use this graph to create a graph of velocity vs. time.

Explanation:

Step1: Recall velocity - position relation

Velocity $v=\frac{\Delta x}{\Delta t}$, which is the slope of the position - time graph.

Step2: Analyze segment A

For segment A from $t = 0\ s$ to $t= 7.5\ s$, the initial position $x_1 = 10\ m$ and the final position $x_2=35\ m$. $\Delta x=x_2 - x_1=35 - 10 = 25\ m$, $\Delta t=7.5\ s$. So $v_A=\frac{25}{7.5}=\frac{10}{3}\approx3.33\ m/s$.

Step3: Analyze segment B

For segment B from $t = 7.5\ s$ to $t = 12.5\ s$, $\Delta x=35 - 35=0\ m$, $\Delta t=12.5 - 7.5 = 5\ s$. So $v_B = 0\ m/s$.

Step4: Analyze segment C

For segment C from $t = 12.5\ s$ to $t = 17.5\ s$, the initial position $x_1 = 35\ m$ and the final position $x_2 = 50\ m$. $\Delta x=x_2 - x_1=50 - 35 = 15\ m$, $\Delta t=17.5 - 12.5=5\ s$. So $v_C=\frac{15}{5}=3\ m/s$.

Step5: Analyze segment D

For segment D from $t = 17.5\ s$ to $t = 20\ s$, the initial position $x_1 = 50\ m$ and the final position $x_2 = 0\ m$. $\Delta x=x_2 - x_1=0 - 50=- 50\ m$, $\Delta t=20 - 17.5 = 2.5\ s$. So $v_D=\frac{-50}{2.5}=-20\ m/s$.

Answer:

The velocity - time graph will have a horizontal line at $v=\frac{10}{3}\ m/s$ from $t = 0\ s$ to $t = 7.5\ s$, a horizontal line at $v = 0\ m/s$ from $t = 7.5\ s$ to $t = 12.5\ s$, a horizontal line at $v = 3\ m/s$ from $t = 12.5\ s$ to $t = 17.5\ s$ and a horizontal line at $v=-20\ m/s$ from $t = 17.5\ s$ to $t = 20\ s$.