Question

use the position - time graph below for q2 - q4. 2. find the average velocity between p and r. 3. at what point(s) is the instantaneous velocity 0? 4. find the instantaneous velocity at 1s.

Explanation:

Step1: Recall average - velocity formula

The average - velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}$, where $\Delta x$ is the change in position and $\Delta t$ is the change in time.
From the graph, the position of point $P$ is $x_P = - 1$ m and the time $t_P=0$ s, and the position of point $R$ is $x_R = 1$ m and the time $t_R = 2$ s.
So, $\Delta x=x_R - x_P=1-(-1)=2$ m and $\Delta t=t_R - t_P=2 - 0 = 2$ s.
$v_{avg}=\frac{\Delta x}{\Delta t}=\frac{2}{2}=1$ m/s.

Step2: Recall instantaneous - velocity concept

The instantaneous velocity at a point on a position - time graph is the slope of the tangent line at that point.
The slope of the tangent line is zero when the graph is horizontal. From the graph, the instantaneous velocity is 0 at point $S$ (because the slope of the tangent line to the curve at point $S$ is 0).

Step3: Find instantaneous velocity at $t = 1$ s

To find the instantaneous velocity at $t = 1$ s, we find the slope of the tangent line at $t = 1$ s.
We can estimate the slope by considering two points close to $t = 1$ s on the curve. Let's take two points: one at $t = 0.5$ s with $x\approx - 0.5$ m and one at $t = 1.5$ s with $x\approx0.5$ m.
The slope $m=\frac{\Delta x}{\Delta t}=\frac{0.5-(-0.5)}{1.5 - 0.5}=\frac{1}{1}=1$ m/s.

Answer:

1. The average velocity between $P$ and $R$ is 1 m/s.
2. The instantaneous velocity is 0 at point $S$.
3. The instantaneous velocity at $t = 1$ s is 1 m/s.