From an aerial photograph, one observes that on a level section of a (multilane) highway, 25% of the vehicles are trucks, 75% are cars, and there are 50 vehicles per mile of highway. The trucks travel at 40 mph and the cars at 60 mph. This highway also has a section with a steep grade on which the speed of the trucks drops to 25 mph and the speed of the cars to 55 mph. No vehicles enter or exit the subject highway sections (except at the ends). The flows on these highwaysections are stationary, which means that the flows on the level section are equal to those on the grade. Determine
(a) the flow of vehicles on the level section;
(b) the density of vehicles on the grade;
(c) the percent of trucks on the grade as seen from an aerial photo; (d) the percent of trucks as seen by a stationary observer on the grade.

(a) Flow rate of vehicles = No of vehicles per mile * Speed

=No of cars per mile * Speed +No of trucks per mile * Speed

= 0.75*50*60 + 0.25*50*40

=2750 vehicles / hour

(b) Let Density of vehicles on grade = x

Density on flat * Speed =Density on grade * Speed

So,( 0.75*50) * 60 + (0.25*50) * 40 = (0.75* x) * 55 + (0.25* x) * 25

So, x= 57.89

So, Density is around 58 Vehicles per Mile.

(c) Percentage of truck by aerial photo = 25%

(d)Percentage of truck bystationary observer on the grade= 25*30/60 * 25/55 =22.73 %

explanation:

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answer:

the formula for amps is watts divided by volts. to use the chart, cover up the a with your finger and use the remaining chart calculation of w divided by v. using our sample panel data, 60 watts divided by 12 volts equals 5 amps.

explanation: