An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in traffic direction) or upstream, or it can be stationary. Figure 2-25shows a uniformly spaced line of cars moving at speed V=25 m/stoward a uniformly spaced line of slow cars moving at speed${\mathit{v}}_{\mathbf{s}}\mathbf{=}\mathbf{5}\mathit{m}\mathbf{/}\mathit{s}$. Assume that each faster car adds length L=12.0 m(car length plus buffer zone) to the line of slow cars when it joins the line, and assume it slows abruptly at the last instant. (a) For what separation distance d between the faster cars does the shock wave remain stationary? If the separation is twice the amount, (b) What is the Speed? (c) What is the Direction (upstream or downstream) of the shock wave?

The length of the car and buffer zone,L=12.0m .

The speed of the faster cars,v=25m/s .

The speed of the slower cars,$v_s=5m/s$ .

Speed is equal to the total distance covered divided by the time taken to cover that distance. For the given cars, find the time using speed, distance, and time relation. Further, calculate the separation distance to keep the shock wave stationary.

Here, the slow car moves distance x and the fast car d+L. When separation distance is doubled, speed, distance, and t can be found. Using that distance and t velocity at which shock wave travels can be calculated.

The direction of movement of the shock wave can be figured out from the value of x, by comparing it with L.

The expression for the speed is given as,

$\mathrm{s}=\frac{\mathrm{Total}\mathrm{distance}}{\mathrm{Total}\mathrm{time}}$ (i)

From the figure, it can be seen that a faster car and a slower car are separated by distance dat t=0

The slow car takes time t to cover the distance L=12.0m . From equation (i), the time t taken to cover the distance L with speed v_s is,

$$\begin{gathered} t=\frac{L}{v_s} \\ =\frac{12\mathrm{m}}{5\mathrm{m}/\mathrm{s}} \\ =2.4\mathrm{s} \end{gathered}$$

In this time the slow car moves distance 12m .

To reach the slow car, the faster car moves distance d+l=vt, for the shock wave to remain stationary, you get d.

$$\begin{gathered} d=vt-L \\ =25.0\mathrm{m}/\mathrm{s}\times 2.4\mathrm{s}-12.0\mathrm{m} \\ =48.0\mathrm{m} \\ \mathrm{Therefore},\mathrm{the}\mathrm{separation}\mathrm{between}\mathrm{the}\mathrm{cars}\mathrm{is}48.0\mathrm{m} \end{gathered}$$

Therefore, the separation between the cars is

When the distance is double that is 96m at t=0 .

After time t slow car moves a distance $x=v_{s}t$in the same time a faster car moves the distance d+x=vt

$$\begin{gathered} t=\frac{x}{v_s} \\ =\frac{d+x}{v} \end{gathered}$$

Rearrange it for xyou get,

$$\begin{gathered} x=\frac{v_s}{v-v_s}\times d \\ =\frac{5.00\mathrm{m}/\mathrm{s}}{25.0\mathrm{m}/\mathrm{s}-5.00\mathrm{m}/\mathrm{s}}\times 96.0\mathrm{m} \\ =24.0\mathrm{m} \end{gathered}$$

To cover this distance, slow car takes the time that is equal to,

$\mathrm{t}=\frac{24.0\mathrm{m}}{5.00\mathrm{m}/\mathrm{s}}$

Slow car pack moves distance,

$$\begin{gathered} ∆\mathrm{x}=\mathrm{x}-\mathrm{L} \\ =24.0\mathrm{m}-12.0\mathrm{m} \end{gathered}$$ downstream.

Speed of the shock wave you get,

$$\begin{gathered} v_{shock}=\frac{∆x}{t} \\ =\frac{12.0\mathrm{m}}{4.8\mathrm{s}} \\ =250\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the speed of shock wave if separation distance doubled is v=2.50 m/s.

The value of x is 24.0m and the value of L is 12m. Therefore, the x is greater than L(x>L) . Hence, the direction of shock wave is downstream.