Question

In: Physics

Let's return to Interstates 80 and 680 in rural Iowa, courtesy of Google Earth. (Open Google...

Let's return to Interstates 80 and 680 in rural Iowa, courtesy of Google Earth. (Open Google Earth using the same file as in the Pre-Lab, Interstate_80_in_Iowa.kmz. (Do it the same way you did in the Chapter 1 lab. If you are using Chrome, there should be a button for this file in the lower left corner of your screen after you download it. If you are using other browsers, this file is probably in your Downloads folder.)
Starting from a complete stop, a car gets on I-80 at the I-80 and I-680 interchange, then drives to Stuart and continues east. After starting to move, the car accelerates over a distance of 1/4 mile, until reaching 55 miles per hour and continuing to Stuart at that constant speed. So between the I-680 interchange and Stuart, the time and distance can be considered to be composed of two time/distance parts:

(link) Interstate_80_in_Iowa.kmz (Google maps shows it is 65.5 miles from I680 to stuart)

time(1): the time taken accelerating from 0 mph to 55 mph during the first 1/4 mile

distance(1): the first 1/4 mile over which the car was accelerating

time(2) the time taken to cover the rest of the distance to Stuart going at the constant speed 55 mph

distance(2): the rest of the distance to Stuart after the 1st 1/4 mile

5.A. What is distance(2), the distance to Staurt after the 1^st ¼ mile? _______________

5.B. What is the time time(1) spent accelerating, in units of hours?___________

5.C. What is the time time(2) spent driving at the constant speed after the 1^st ¼ mile to Stuart? ____________

5.D. - What percentage of the total distance of the trip (distance(1) + distance(2)) between I-680 interchange and Stuart was spent accelerating?

5.E. What percentage of the total time of this trip (time(1) + time(2)) was spent accelerating?

Expert Solution

The acceleration of the car while covering the distance of 1/4 mile can be found by using the kinematical equation:

where v is its velocity after covering the 1/4 mile,a is its acceleration and s is 1/4 mile.

Also u=0 m/s, v=55 mi/h=55*0.45 m/s=24.8 m/s.

Also 1/4 mile=1/4*1609 m=402.3 m.

Therefore substituting these values in the above equation we obtain:

i.e a=615.04/804.5=0.76m/s² in S.I.units.

The time taken by the car to cover the first 1/4 mile is given by the equation v=u+at where u=0 m/s, v=55 mi/h=24.8 m/s and a=0.76 m/s².

Therefore t₁=v/a=24.8/0.76=32.6 s.

A.The distance to Stuart after the first 1/4 mile=65.5 mile-1/4mile=65.25 mile=1.05*10⁵ m.

B.The time spent in accelerating=32.6 s=32.6/3600=0.009 hours.

C.The time spent in driving at constant speed=62.25 mi/55 mi/h=1.13 hours=4075 sec.

D.Percentage of distance spent in acceleration=1/4 mi/1680 mi*100=0.015%.

E.Percentage of time spent in accelerating=32.6s/(32.6+4075)s=0.8%.

genius_generous answered 5 months ago

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