Question

1. a bus makes a trip according to the position - time graph shown in the drawing. what is average velocity of the bus during each of the segments labeled a, b, and c? express yo answers in km/h.

a person who walks for exercise produces the position - time graph given with (a) without making any calculations, decide which segments of the graph (a indicate positive, negative, and zero average velocities.
(b) calculate the average velocity for each segment to verify your answers

Explanation:

Step1: Recall average - velocity formula

The average velocity formula is $v_{avg}=\frac{\Delta x}{\Delta t}$, where $\Delta x = x_f - x_i$ and $\Delta t=t_f - t_i$.

Step2: Analyze segment A of the first graph

For segment A of the first graph: $x_i = 40\ km$, $x_f = 10\ km$, $t_i = 0\ h$, $t_f = 1.5\ h$. Then $\Delta x=10 - 40=- 30\ km$ and $\Delta t = 1.5-0 = 1.5\ h$. So $v_{avg,A}=\frac{-30}{1.5}=-20\ km/h$.

Step3: Analyze segment B of the first graph

For segment B of the first graph: $x_i = 10\ km$, $x_f = 20\ km$, $t_i = 1.5\ h$, $t_f = 2.5\ h$. Then $\Delta x=20 - 10 = 10\ km$ and $\Delta t=2.5 - 1.5 = 1\ h$. So $v_{avg,B}=\frac{10}{1}=10\ km/h$.

Step4: Analyze segment C of the first graph

For segment C of the first graph: $x_i = 20\ km$, $x_f = 40\ km$, $t_i = 2.5\ h$, $t_f = 3.0\ h$. Then $\Delta x=40 - 20 = 20\ km$ and $\Delta t=3.0 - 2.5 = 0.5\ h$. So $v_{avg,C}=\frac{20}{0.5}=40\ km/h$.

For the second - graph part (a)

A segment with a positive slope on a position - time graph has a positive average velocity, a segment with a negative slope has a negative average velocity, and a horizontal segment has a zero average velocity. So for the second graph: Segment A has a positive average velocity, segment B has a negative average velocity, and segment D has a zero average velocity.

For the second - graph part (b)

Let's assume for segment A: $x_i = 0$, $x_f = 1.25$, $t_i = 0$, $t_f = 0.25$. Then $\Delta x=1.25-0 = 1.25$ and $\Delta t=0.25 - 0 = 0.25$. So $v_{avg,A}=\frac{1.25}{0.25}=5$ (units assumed to be km/h if position is in km and time in h).
For segment B: $x_i = 1.25$, $x_f = 0.5$, $t_i = 0.25$, $t_f = 0.5$. Then $\Delta x=0.5 - 1.25=-0.75$ and $\Delta t=0.5 - 0.25 = 0.25$. So $v_{avg,B}=\frac{-0.75}{0.25}=-3$ (km/h).
For segment C: assume $x_i = 0.5$, $x_f = 0.75$, $t_i = 0.5$, $t_f = 1.0$. Then $\Delta x=0.75 - 0.5 = 0.25$ and $\Delta t=1.0 - 0.5 = 0.5$. So $v_{avg,C}=\frac{0.25}{0.5}=0.5$ (km/h).
For segment D: $x_i = 0.75$, $x_f = 0.75$, $t_i = 1.0$, $t_f = 1.5$. Then $\Delta x=0.75 - 0.75 = 0$ and $\Delta t=1.5 - 1.0 = 0.5$. So $v_{avg,D}=\frac{0}{0.5}=0$ (km/h).

Answer:

For the first graph:
$v_{avg,A}=-20\ km/h$, $v_{avg,B}=10\ km/h$, $v_{avg,C}=40\ km/h$
For the second graph part (a): Segment A has positive average velocity, segment B has negative average velocity, segment D has zero average velocity.
For the second graph part (b): $v_{avg,A}=5\ km/h$, $v_{avg,B}=-3\ km/h$, $v_{avg,C}=0.5\ km/h$, $v_{avg,D}=0\ km/h$