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Assuming-it-rained-at-a-constant-rate-and-the-rain-fell-at-angle-to-the-ground-see-diagram-determine-if-walking-or-running-causes-you-to-get-more-less-wet-or-of-it-makes-no-difference-for-1-

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Question Number 11902 by FilupS last updated on 04/Apr/17
Assuming it rained at a constant rate, and the rain fell at angle θ to the ground (see diagram), determine if walking or running causes you to get more/less wet, or of it makes no difference for: 1. θ=90° (downwards) 2. θ<90° (the rain is moving on the same direction as you) 3. θ>90° (the rain is moving/blowing into you)
$$\mathrm{Assuming}\:\mathrm{it}\:\mathrm{rained}\:\mathrm{at}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{rate}, \\ $$$$\mathrm{and}\:\mathrm{the}\:\mathrm{rain}\:\mathrm{fell}\:\mathrm{at}\:\mathrm{angle}\:\theta\:\mathrm{to}\:\mathrm{the}\:\mathrm{ground} \\ $$$$\left(\mathrm{see}\:\mathrm{diagram}\right),\:\mathrm{determine}\:\mathrm{if}\:\mathrm{walking}\:\mathrm{or} \\ $$$$\mathrm{running}\:\mathrm{causes}\:\mathrm{you}\:\mathrm{to}\:\mathrm{get}\:\mathrm{more}/\mathrm{less}\:\mathrm{wet}, \\ $$$$\mathrm{or}\:\mathrm{of}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{difference}\:\mathrm{for}: \\ $$$$\: \\ $$$$\mathrm{1}.\:\:\:\theta=\mathrm{90}°\:\:\left(\mathrm{downwards}\right) \\ $$$$\mathrm{2}.\:\theta<\mathrm{90}°\:\:\left(\mathrm{the}\:\mathrm{rain}\:\mathrm{is}\:\mathrm{moving}\:\mathrm{on}\:\mathrm{the}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{same}\:\mathrm{direction}\:\mathrm{as}\:\mathrm{you}\right) \\ $$$$\mathrm{3}.\:\theta>\mathrm{90}°\:\:\left(\mathrm{the}\:\mathrm{rain}\:\mathrm{is}\:\mathrm{moving}/\mathrm{blowing}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\mathrm{into}\:\mathrm{you}\right) \\ $$
Commented by FilupS last updated on 04/Apr/17
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 04/Apr/17
we can suppose that the human budy like a rectangle that its dimentions are almost about:B=25,W=65,H=180,in centimeters. 1.θ<90 or θ>90 are the same.in this case wet area=65×180#11700cm^2 area above sholders=(65−25)×25=1000 added area for man=2πR_1 (r_1 +2h_1 )+πR_2 (r_2 +2h_2 )= =2×3.14×12(5+2×4)+1×3.14×4(2+2×1.5)=981+63=1043 added area for lady=2π×16(8+3×5)+2×π×10(5+3×4)=2311+1068=3379 total for man ,face to rain= 11700−1000+63=10763cm^2 total for man, back to rain= 11700−1000+981=11681cm^2 total for lady,face to rain= 11700−1000+1068=11768cm^2 total for lady ,back to rain= 11700−1000+2311=13011cm^2 2.θ=90,in this case only up side of rectangle opposite to rain.so: the wet area =25×65#1625cm^2 total for man=1625+(1/2)×63+(1/2)×981#2138cm^2 total for lady=1625+(1/2)×2311+(1/2)×1068#3315cm^2 Now we know that the max of wet area in case of:θ>90^° and for lady that run back to rain.I think running or walking are the same .excuse me,I think we should calculating areas with much more of accuracy.
$${we}\:{can}\:{suppose}\:{that}\:{the}\:{human}\:{budy} \\ $$$${like}\:{a}\:{rectangle}\:{that}\:{its}\:{dimentions}\:{are} \\ $$$${almost}\:{about}:{B}=\mathrm{25},{W}=\mathrm{65},{H}=\mathrm{180},{in} \\ $$$${centimeters}. \\ $$$$\mathrm{1}.\theta<\mathrm{90}\:{or}\:\theta>\mathrm{90}\:{are}\:{the}\:{same}.{in}\:{this} \\ $$$${case}\:{wet}\:{area}=\mathrm{65}×\mathrm{180}#\mathrm{11700}{cm}^{\mathrm{2}} \\ $$$${area}\:{above}\:{sholders}=\left(\mathrm{65}−\mathrm{25}\right)×\mathrm{25}=\mathrm{1000} \\ $$$${added}\:{area}\:{for}\:{man}=\mathrm{2}\pi{R}_{\mathrm{1}} \left({r}_{\mathrm{1}} +\mathrm{2}{h}_{\mathrm{1}} \right)+\pi{R}_{\mathrm{2}} \left({r}_{\mathrm{2}} +\mathrm{2}{h}_{\mathrm{2}} \right)= \\ $$$$=\mathrm{2}×\mathrm{3}.\mathrm{14}×\mathrm{12}\left(\mathrm{5}+\mathrm{2}×\mathrm{4}\right)+\mathrm{1}×\mathrm{3}.\mathrm{14}×\mathrm{4}\left(\mathrm{2}+\mathrm{2}×\mathrm{1}.\mathrm{5}\right)=\mathrm{981}+\mathrm{63}=\mathrm{1043} \\ $$$${added}\:{area}\:{for}\:{lady}=\mathrm{2}\pi×\mathrm{16}\left(\mathrm{8}+\mathrm{3}×\mathrm{5}\right)+\mathrm{2}×\pi×\mathrm{10}\left(\mathrm{5}+\mathrm{3}×\mathrm{4}\right)=\mathrm{2311}+\mathrm{1068}=\mathrm{3379} \\ $$$${total}\:{for}\:{man}\:,{face}\:{to}\:{rain}= \\ $$$$\mathrm{11700}−\mathrm{1000}+\mathrm{63}=\mathrm{10763}{cm}^{\mathrm{2}} \\ $$$${total}\:{for}\:{man},\:{back}\:{to}\:{rain}= \\ $$$$\mathrm{11700}−\mathrm{1000}+\mathrm{981}=\mathrm{11681}{cm}^{\mathrm{2}} \\ $$$${total}\:{for}\:{lady},{face}\:{to}\:{rain}= \\ $$$$\mathrm{11700}−\mathrm{1000}+\mathrm{1068}=\mathrm{11768}{cm}^{\mathrm{2}} \\ $$$${total}\:{for}\:{lady}\:,{back}\:{to}\:{rain}= \\ $$$$\mathrm{11700}−\mathrm{1000}+\mathrm{2311}=\mathrm{13011}{cm}^{\mathrm{2}} \\ $$$$\mathrm{2}.\theta=\mathrm{90},{in}\:{this}\:{case}\:{only}\:{up}\:{side}\:{of} \\ $$$${rectangle}\:{opposite}\:{to}\:{rain}.{so}: \\ $$$${the}\:{wet}\:{area}\:=\mathrm{25}×\mathrm{65}#\mathrm{1625}{cm}^{\mathrm{2}} \\ $$$${total}\:{for}\:{man}=\mathrm{1625}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{63}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{981}#\mathrm{2138}{cm}^{\mathrm{2}} \\ $$$${total}\:{for}\:{lady}=\mathrm{1625}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2311}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{1068}#\mathrm{3315}{cm}^{\mathrm{2}} \\ $$$${Now}\:{we}\:{know}\:{that}\:{the}\:{max}\:{of}\:{wet}\:{area} \\ $$$${in}\:{case}\:{of}:\theta>\mathrm{90}^{°} {and}\:{for}\:{lady}\:{that}\:{run} \\ $$$${back}\:{to}\:{rain}.{I}\:{think}\:{running}\:{or}\:{walking}\:{are}\:{the}\:{same}\:.{excuse}\:{me},{I}\:{think}\:{we}\:{should}\:{calculating}\:{areas} \\ $$$${with}\:{much}\:{more}\:{of}\:{accuracy}. \\ $$

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