UNIT ANALYSIS Word Problem?

Question

At 65 miles per hour, how many car lengths of spacing should you leave between you and a car ahead of you for a 3-second margin of safety? In other words, how many car lengths would you travel in those 3 seconds? Suppose one car length is 20 feet. USE UNIT ANALYSIS to answer the question. Round your answer to the nearest tenth. Show all workd and explain how you arrived at your conclusion? I just started learning this .. could someone walk me through this step by step .. I would greatly appreciate it!

Answer 1

65mi/hr equals 95.33ft/sec
the travel distance: 95.33ft/s*3s=285.99ft
if 1 car equals 20ft then 285.99ft equals 14 cars and 6ft

Answer 2

With unit analysis, you are basically converting and cancelling out units from denominators and numerators. So a car traveling at 65 miles /hour wile travel at 65 miles /hr x 1 hr/3600 secs = 0.0180 miles / sec. In the 3 seconds, the car would travel 0.054 miles. Then coverting miles to ft with 1 mile = 5280 ft = 286 ft which covers 286/20 = 14.2 car length.

or

65 miles / hour x 1 hour/3600 secs x 5280 ft /mile x car length / 20 ft x 3 secs = 14.3 car length when you have crossed out units from denominators and numerators.

Answer 3

You are going 65 miles/hr = 65*5280ft/hr = 343,200 ft/hr

1 hour = 3600 secs

343,200 ft/hr = 343,200ft/3600secs = 95.333 ft/sec

so in 3 secs you would travel approximately 286 feet.

At 20 feet for a car you should leave 286/20 = 14.3 car lengths

Answer 4

The roots are all on the unit circle, yet so a techniques i will't really educate it. i will educate that the variety of roots not on the unit circle is a numerous of four, so as that sort is both 0, 4, or 8, that means that both all 10, or 6, or 2 roots are on the unit circle. right that is my artwork so a techniques: enable f(z) be the given polynomial. evaluate g(z) = -f(iz) = 11z^10 +10z^9 + 10z + 11. when you consider that z->iz and z->-z both map the unit circle to itself, the variety of roots of g(z) on the unit circle is an same because the variety of roots of f(z). So we would to boot do the project for g(z) extremely than f(z). Now, g(z) has authentic coefficients, so if z' is a root of g(z) so is conj(z'), the conjugate of z'. also, that's common to study that g(z) has no authentic roots, so z' and conj(z') could be diverse. also, because the coefficients of g(z) are palindromic, that's self-reciprocal. this suggests that if z' is a root then so is a million/z'. Now a complicated style, z', is on the unit circle iff conj(z') = a million/z'. subsequently, if z' is a root of g(z) not on the unit circle then the 4 diverse complicated numbers z', conj(z'), a million/z', conj(a million/z') = a million/conj(z') are all roots of g(z) not on the unit circle, 2 interior and a pair of outside the unit circle. This proves that the variety of roots of g(z) not on the unit circle is a numerous of four. And subsequently a minimum of two zeros of g(z) are on the unit circle. to end, one needs to educate that each and each and every body the reciprocals and conjugates of a few 5 roots, would nicely be paired, like this: pair one: z1, a million/z1 = conj(z1); pair 2: z2, a million/z2 = conj(z2), ..., pair 5: z5, a million/z5 = conj(z5). possibly someone sees what i'm lacking and would end. yet for now that is all I have. ---- ---- ---- EDIT: @kakos: to ascertain that g has no authentic zeros evaluate it as a real polynomial of a real variable. in case you plot it you'll see the graph lies completely above the x-axis. For a suitable evidence: through Descartes rule of indicators g has both 2 or 0 authentic roots, and if 2 they are both adverse. yet when you consider that g is palindromic if z' is a root so is a million/z'. this suggests if there are authentic roots one is less than -a million and one is between -a million and 0. yet g(-a million) = 2 and g'(z)<0 for z<-a million, so g(z)>2 for all z<-a million. subsequently g has no 0 less than -a million. subsequently the variety of authentic roots is 0. --- --- --- EDIT: @jaz_will; magnificent interest! you deserve BA.

Answer 5

(65 mi/hr)(1 hr/3600 sec)(3 sec)(5,280ft/1 mi)(1 car length/20 ft.) = 14.3 car lengths

(Good luck doing this on any highway!)