Question

Consider an 8-car caravan, where the propagation speed is 100 km/hour, each car takes 1 minute to pass a toll both. The caravan starts in front of toll booth A, goes through toll booth B, and ends after passing toll booth C. Let dAB and dBC be the distance between A-B, and B-C. a) Suppose dAB = dBC = 10 km. What is the end-to-end delay if the caravan travels together (i.e., the first car must wait for the last car after passing each toll booth)?b) Repeat a), but assume the cars travel separately (i.e., not waiting for each other). c)Repeat a) and b), but suppose dAB = dBC = 100 km. d) Still suppose dAB = dBC = 100 km. Suppose toll booth B takes 10 minute to pass each car (A and C still takes 1 minute per car). Where is the first car when the second car passes B? e) Under the assumption of d), what is the maximum value of dBC such that the first car has passed C when the second car passes B?

Answer 1

Answer:

Part a :  If the distance is 10 km each, the total end to end delay if caravan travels together is 36 mins.

Part b: If the distance is 10 km each, the total end to end delay if cars travel separately is 120 mins.

Part c: If the distance is 100 km each, the total end to end delay if caravan travels together is 144 mins and 984 mins,  if cars travel separately.

Part d: car 1 is 1.67 km ahead of car 2 when car 2 passes toll B.

Part e: The maximum value of dBC such that the first car has passed C when the second car passes B is 98. 3 km.

Step-by-step explanation:

Part a

Here dAB=dBC=10km

Propagation delay is given as

                                        $d_{prop}=\frac{distance travelled}{propagation speed}\\d_{prop}=\frac{dAB+dBC }{propagation speed}\\d_{prop}=\frac{10+10}{100}\\d_{prop}=0.2 hr\\d_{prop}=0.2 \times 60 \\d_{prop}=12 mins$

Translation delay at toll both is given as

                                $d_{trans}=n_{toll} \times n_{cars} \times delay_{unit}\\d_{trans}=3 \times 8 \times 1 min\\d_{trans}=24 min$

Total End to End delay is given as

                              $d_{total}=d_{prop}+d_{trans}\\d_{total}=12+24 min\\d_{total}=36 min$

So the total end to end delay if caravan travels together is 36 mins.

Part b

For this part, Propagation delay will remain same, However the Translation delay is given as

                               $(d_{trans})_{n_c=1}=n_{toll} \times delay_{unit}\\(d_{trans})_{n_c=1}=3 \times 1 min\\(d_{trans})_{n_c=1}=3 min$

So total delay for one car is given as

$(d_{total})_{n_c=1}=d_{prop}+(d_{trans})_{n_c=1}\\(d_{total})_{n_c=1}=12+3 min\\(d_{total})_{n_c=1}=15 min$

Total delay for 8 cars is given as

$d_{total}=n_{car} \times (d_{total})_{n_c=1}\\d_{total}=8 \times 15\\d_{total}=120 mins$

So the total end to end delay if cars travel separately is 120 mins.

Part c

Here dAB=dBC=100km

When Caravan travels together

Propagation delay is given as

                                      $d_{prop}=\frac{distance travelled}{propagation speed}\\d_{prop}=\frac{dAB+dBC }{propagation speed}\\d_{prop}=\frac{100+100}{100}\\d_{prop}=2 hr\\d_{prop}=2 \times 60 \\d_{prop}=120 mins$

Translation delay at toll both is given as

                                $d_{trans}=n_{toll} \times n_{cars} \times delay_{unit}\\d_{trans}=3 \times 8 \times 1 min\\d_{trans}=24 min$

Total End to End delay is given as

                              $d_{total}=d_{prop}+d_{trans}\\d_{total}=120+24 min\\d_{total}=144 min$

So the total end to end delay if caravan travels together is 144 mins.

When Cars travel separately

For this part, Propagation delay will remain same, However the Translation delay is given as

                               $(d_{trans})_{n_c=1}=n_{toll} \times delay_{unit}\\(d_{trans})_{n_c=1}=3 \times 1 min\\(d_{trans})_{n_c=1}=3 min$

So total delay for one car is given as

$(d_{total})_{n_c=1}=d_{prop}+(d_{trans})_{n_c=1}\\(d_{total})_{n_c=1}=120+3 min\\(d_{total})_{n_c=1}=123 min$

Total delay for 8 cars is given as

$d_{total}=n_{car} \times (d_{total})_{n_c=1}\\d_{total}=8 \times 123\\d_{total}=984 mins$

So the total end to end delay if cars travel separately is 984 mins.

Part d

When the car 1 reaches toll A, the car 2 is 1 min behind.

When car 1 reaches toll B, the car 2 is 1 min behind.

The car 1 leaves toll B after 10 mins.

The car 2 leaves toll B after 11 mins.

So the total delay difference is 11 mins-10 mins

Distance traveled in given time is given as

$d_{travel}=t \times v\\d_{travel}=(1/60) hr \times 100 km/hr\\d_{travel}=1.67 km$

So car 1 is 1.67 km ahead of car 2 when car 2 passes toll B.

Part e

The maximum value of dBC is given as

$d_{BC_{max}} =d_{BC}-d_{travel}\\d_{BC_{max}} =100-1.67\\d_{BC_{max}} =98.3 km$

The maximum value of dBC such that the first car has passed C when the second car passes B is 98. 3 km.