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Pmand92

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An abrupt slowdown in concentrated traffic can travel as a pulse, termed a shock wave, along the line of cars, either downstream (in the traffic direction) or upstream, or it can be stationary. figure below shows a uniformly spaced line of cars moving at speed v = 26.0 m/s toward a uniformly spaced line of slow cars moving at speed v_s = 5.20 m/s. Assume that each faster car adds length L = 15.0 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 that amount, what are the (b) speed and (c) direction (upstream or downstream) of the shock wave?

2. Homework Equations :

Incoming Flux (Fast zone): (1 car/d) * (26 m/s)

d=Unknown distance

Outgoing Flux (Slow zone): (1 car/15 m)* (5.20 m/s)

1) Δt= Δr/ΔV(avg)

2) r=V(avg)*t

3. The Attempt at a Solution :

a) I have used the conversion about the outgoing and incoming flux equations and set them equal to each other to find d, in meters. I got 75.0 meters, which was not correct in WileyPlus. Then I tried for this same part using equation 1: Δt= 15.0 m/5.20 m/s, which gave me 2.8 seconds for each car to join the "slow zone" every 2.8 seconds. Then I used the 2nd equations with the new found time: r=26.0 m/s*2.8 seconds, which gave me 72.8 m. This again was not correct in WileyPlus.

b) I tried to follow an answer that was posted on YahooAnswers about this part, except replacing their variables with mine:

"If the cars are separated by twice this distance, then 12m get added to the end of the shock zone every 4.8s (120/25). So that comes out to 2.5m/s (12/4.8). Thus the shockwave advances down the roadway at 2.5m/s (5-2.5)". After I tried this with my variables, again the answer was not correct in WileyPlus.