Chapter 6, Problem 1P

To determine

(a)

Time mean speed.

Answer to Problem 1P

The time mean speed is $\overline{{\text{u}}_{\text{t}}}\text{ = 47}\text{.19}\frac{\text{miles}}{\text{hour}}.$

Explanation of Solution

Given:

Observers stationed at two sections XX and YY, 500 ft apart on a highway, recording the time of vehicles on their arrival as shown in the accompanying table

The total time of observation at XX was 15 sec.

In order to calculate time mean speed, we need to figure out traveling time of each vehicle which is the difference of time taken by a vehicle in travelling from XX to YY section between the two sections.

Calculation:

Vehicles	Section XX	Section YY	Travelling time of vehicle,   $t_i$
A	$T_0$	$T_0+7.85\sec$	$\left(7.85-0\right)=7.85$   $\mathrm{Sec}$
B	$T_0+3\sec$	$T_0+9.18\sec$	$\left(9.18-3.00\right)=6.18$   $\mathrm{Sec}$
C	$T_0+6\sec$	$T_0+12.36\sec$	$\left(12.36-6.0\right)=6.36$   $\mathrm{Sec}$
D	$T_0+12\sec$	$T_0+21.74\sec$	$\left(21.74-12.0\right)=9.74$   $\mathrm{Sec}$

We have the following formula for the time mean speed:

  $\overline{{\text{u}}_{\text{t}}}\text{ = }\frac{1}{\text{n}}{\displaystyle {\sum }_{\text{i = 1}}^{\text{n}}{\text{u}}_{\text{i}}}$

Where, ${\text{u}}_{\text{i}}$ is the speed of the i^th vehicle.

n is the number of the vehicles passing.

Now, the speed of the individual vehicles is as follows:

Speed of vehicle A:

  $\begin{array}{l}{\text{u}}_{\text{1}}\text{=}{\text{u}}_{\text{A}}=\frac{\text{L}}{{\text{t}}_{\text{A}}}\\ {\text{u}}_{\text{1}}\text{=}{\text{u}}_{\text{A}}=\frac{500\text{ft}}{7.58\sec }\\ {\text{u}}_{\text{1}}\text{=}{\text{u}}_{\text{A}}=65.963\frac{\text{ft}}{\sec }.\end{array}$

Speed of vehicle B:

  $\begin{array}{l}{\text{u}}_{\text{2}}\text{=}{\text{u}}_{\text{B}}=\frac{\text{L}}{{\text{t}}_{\text{B}}}\\ {\text{u}}_{\text{2}}\text{=}{\text{u}}_{\text{B}}=\frac{500\text{ft}}{6.18\sec }\\ {\text{u}}_{\text{2}}\text{=}{\text{u}}_{\text{B}}=80.91\frac{\text{ft}}{\sec }.\end{array}$

Speed of vehicle C:

  $\begin{array}{l}{\text{u}}_{\text{3}}={\text{u}}_{\text{C}}=\frac{\text{L}}{{\text{t}}_{\text{C}}}\\ {\text{u}}_{\text{3}}={\text{u}}_{\text{C}}=\frac{500\text{ft}}{7.58\sec }\\ {\text{u}}_{\text{3}}={\text{u}}_{\text{C}}=78.62\frac{\text{ft}}{\sec }.\end{array}$

Speed of vehicle D:

  $\begin{array}{l}{\text{u}}_{\text{4}}={\text{u}}_{\text{D}}=\frac{\text{L}}{{\text{t}}_{\text{D}}}\\ {\text{u}}_{\text{4}}={\text{u}}_{\text{D}}=\frac{500\text{ft}}{9.74\sec }\\ {\text{u}}_{\text{4}}={\text{u}}_{\text{D}}=51.335\frac{\text{ft}}{\sec }.\end{array}$

Now, the time mean speed can be calculated by substituting the values in the following formula:

  $\begin{array}{l}\overline{{\text{u}}_{\text{t}}}\text{ = }\frac{1}{\text{n}}{\displaystyle {\sum }_{\text{i = 1}}^{\text{n}}{\text{u}}_{\text{i}}}\\ \overline{{\text{u}}_{\text{t}}}\text{ = }\frac{1}{4}\left[{\text{u}}_{\text{1}}{\text{+u}}_{\text{2}}{\text{+u}}_{\text{3}}{\text{+u}}_{\text{4}}\right]\\ \overline{{\text{u}}_{\text{t}}}\text{ = }\frac{1}{4}\left[65.963\frac{\text{ft}}{\sec }\text{+}80.91\frac{\text{ft}}{\sec }\text{+}78.62\frac{\text{ft}}{\sec }\text{+}51.335\frac{\text{ft}}{\sec }\right]\\ \overline{{\text{u}}_{\text{t}}}\text{ = }\frac{1}{4}\left[276.828\frac{\text{ft}}{\sec }\right]\\ \overline{{\text{u}}_{\text{t}}}\text{ = 69}\text{.207}\frac{\text{ft}}{\sec }\\ \overline{{\text{u}}_{\text{t}}}\text{ = 69}\text{.207}\frac{\text{ft}}{\sec }\times \frac{\text{0}\text{.000189394}\text{miles}\text{60}\times \text{60}}{\text{hour}}\times \frac{\text{sec}}{\text{ft}}\because 1\text{ft }\text{= 0}\text{.000189394}\text{miles}\\ \overline{{\text{u}}_{\text{t}}}\text{ = 69}\text{.207}\times \frac{\text{0}\text{.000189394}\times \text{3600}\text{miles}}{\text{hour}}\\ \overline{{\text{u}}_{\text{t}}}\text{ = 69}\text{.207}\times \frac{\text{0}\text{.681818}\text{miles}}{\text{hour}}\\ \overline{{\text{u}}_{\text{t}}}\text{ = 47}\text{.19}\frac{\text{miles}}{\text{hour}}.\end{array}$

Conclusion:

Therefore, the time mean speed is $\overline{{\text{u}}_{\text{t}}}\text{ = 47}\text{.19}\frac{\text{miles}}{\text{hour}}.$

To determine

(b)

Space mean speed.

Answer to Problem 1P

The time mean speed is $\overline{{\text{u}}_{\text{t}}}\text{ = 45}\text{.668}\frac{\text{miles}}{\text{hour}}.$

Explanation of Solution

Given:

Observers stationed at two sections XX and YY, 500 ft apart on a highway, recording the time of vehicles on their arrivalas shown in the accompanying table.

The total time of observation at XX was 15 sec.

Calculation:

We have the following formula for finding out the space mean speed.

  $\overline{{\text{u}}_{\text{S}}}\text{ = }\frac{\text{L}\times \text{n}}{{\displaystyle {\sum }_{\text{i = 1}}^{\text{n}}{\text{t}}_{\text{i}}}}$

Where, ${\text{t}}_{\text{i}}$ is the time that i^thvehicle takes to travel from XX to YY at the speed of ${\text{u}}_{\text{i}}$.

Substitute the values, we have:

  $\begin{array}{l}\overline{{\text{u}}_{\text{S}}}\text{ = }\frac{\text{L}\times \text{n}}{{\displaystyle {\sum }_{\text{i = 1}}^{\text{n}}{\text{t}}_{\text{i}}}}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = }\frac{500\text{ ft}\times 4}{{\displaystyle {\sum }_{\text{i = 1}}^4{\text{t}}_{\text{i}}}}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = }\frac{500\text{ ft}\times 4}{\left[7.58\text{sec}\text{+}\text{6}\text{.18}\text{sec}\text{+ 6}\text{.36sec}\text{+}\text{9}\text{.74sec}\right]}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = }\frac{500\text{ ft}\times 4}{\left[\text{29}\text{.86}\text{sec}\right]}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = }\frac{\text{66}\text{.98 ft}}{\text{sec}}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = }\frac{\text{66}\text{.98 ft}}{\text{sec}}\times \frac{\text{0}\text{.000189394}\text{miles}\text{60}\times \text{60}}{\text{hour}}\times \frac{\text{sec}}{\text{ft}}.\left[\because 1\text{ft }\text{= 0}\text{.000189394}\text{miles}\right]\\ \overline{{\text{u}}_{\text{S}}}\text{= 66}\text{.98}\times \frac{\text{0}\text{.000189394}\times \text{3600}\text{miles}}{\text{hour}}.\\ \overline{{\text{u}}_{\text{S}}}\text{ = 66}\text{.98}\times \frac{\text{0}\text{.681818}\text{miles}}{\text{hour}}.\\ \overline{{\text{u}}_{\text{t}}}\text{ = 45}\text{.668}\frac{\text{miles}}{\text{hour}}.\end{array}$

Conclusion:

Therefore, the time mean speed is $\overline{{\text{u}}_{\text{t}}}\text{ = 45}\text{.668}\frac{\text{miles}}{\text{hour}}.$

To determine

(c)

Flow at section XX.

Answer to Problem 1P

The flow at section XX is as follows $\frac{960\text{Vehicles}}{\text{hour}}$.

Explanation of Solution

Given:

Observers stationed at two sections XX and YY, 500 ft apart on a highway, recording the time of vehicles on their arrival as shown in the accompanying table.

The total time of observation at XX was 15 sec.

Calculation:

We have the following formula for finding out the flow at XX section.

  ${\text{q}}_{\text{xx}}\text{= }\frac{\text{n}}{{\text{T}}_{\text{xx}}}$

Where, ${\text{T}}_{\text{xx}}$ is the total time of observation at XXsection.

Substitute the values, we have:

  $\begin{array}{l}{\text{q}}_{\text{xx}}\text{= }\frac{\text{4}\text{Vehicles}}{\text{15}\text{sec}}\times \frac{3600\sec }{\text{hour}}\\ {\text{q}}_{\text{xx}}\text{= }\frac{\text{4}\text{Vehicles}\times \text{3600}}{\text{15}\text{hour}}\\ {\text{q}}_{\text{xx}}\text{= }\frac{\text{14}400\text{Vehicles}}{\text{15}\text{hour}}\\ {\text{q}}_{\text{xx}}\text{= }\frac{960\text{Vehicles}}{\text{hour}}\end{array}$

Conclusion:

Therefore, the flow at section XX is as follows $\frac{960\text{Vehicles}}{\text{hour}}$.