Question:Compute your average velocity in following two cases: (a) You walk 73.2 mat a speed of 1.22 m/sand then run 73.2at a speed of 3.05m/salong a straight track. (b) You walk for1 minat a speed of 1.22m/sand then run for1 min at3.05m/salong a straight track. (c) Graph x versus t for both cases and indicate how the average velocity is found on the graph.

In the first case, the distance covered while walking,x₁ =73.2 m .

In the first case, the speed while walking,v₁ =1.22 m/s .

In the first case, the distance covered while running,x₂ =73.2 m .

In the first case, the speed while running, v₂ =3.05 m/s .

In the second case, the time for walking,t₁ =1 min .

In the second case, the speed while walking,v₁ =1.22 m/s .

In the second case, the time for running, t₂ =1 min .

In the second case, the velocity while running, v₂ =3.05 m/s .

Average velocity is the ratio of the total displacement to the total time taken to occur that displacement.

The expression for the average velocity is as follows:

$v_{avg}=\frac{∆x}{∆t}$ (i)

Here,$∆x$ is the displacement, and $∆t$is the time interval.

The total time taken to cover the whole distance is calculated as follows:

$$\begin{gathered} ∆t=\frac{x_1}{v_1}+\frac{x_2}{v_2} \\ =\frac{73.2m}{1.22m/s}+\frac{73.2m}{3.05m/s} \\ =60s+24s \\ =84s \end{gathered}$$

The total distance covered is calculated as follows:

$$\begin{gathered} ∆x=x_1+x_2 \\ =73.2+73.2 \\ =146.4m \end{gathered}$$

Now, the average velocity is as follows:

role="math" localid="1651662453157" $$\begin{gathered} ∆v_{avg}=\frac{∆x}{∆t} \\ =\frac{146.4m}{84s} \\ =1.74m/s \end{gathered}$$

Thus, the average velocity in the first case is 1.74 m/s.

In this case, the total distance covered is calculated as follows:

$$\begin{gathered} ∆x=v_1\times t_1+v_2\times t_2 \\ =1.22m/s\times 60s+3.05m/s\times 60s \\ =73.2+183 \\ =256.2m \end{gathered}$$

The total time interval is as follows:

$$\begin{gathered} ∆t=60+60 \\ =120s \end{gathered}$$

So, the average velocity is as follows:

$$\begin{gathered} ∆v_{avg}=\frac{∆x}{∆t} \\ =\frac{256.2m}{120s} \\ =2.135m/s \end{gathered}$$

Therefore, the average velocity in the second case is 2.135 m/s.

You can find the average velocity using a graph ofx versus t.

Case (a)

The graph shows two different displacement paths with a blue line. The first one is the displacement while walking, and the second one is the displacement while running. If you take the slope of each line, you will get the velocity of walking and running. If you take the slope of the line connecting the initial displacement to the final displacement (black line), you will get the average velocity.

Case (b)

The graph shows two displacements. The first one is walking for,and the second one is running for. The slope of each line will give us the velocity of walking and running.If you take the slope of the line connecting the initial displacement to the final displacement (black line), you will get the average velocity.