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ieatpi
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A space probe is shot upward from the earth. If air resistance is disregarded, a differential equation for the velocity after burnout is v dv/dt = ky^2, where y is the distance from the center of the earth and k is a positive constant. If y0 is the distance from the center of the earth at burnout and v0 is the corresponding velocity, express v as a function of y.
 one year ago
 one year ago
ieatpi Group Title
A space probe is shot upward from the earth. If air resistance is disregarded, a differential equation for the velocity after burnout is v dv/dt = ky^2, where y is the distance from the center of the earth and k is a positive constant. If y0 is the distance from the center of the earth at burnout and v0 is the corresponding velocity, express v as a function of y.
 one year ago
 one year ago

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hal_stirrup Group TitleBest ResponseYou've already chosen the best response.0
write the equation again am not sure what am reading is right.
 one year ago

ieatpi Group TitleBest ResponseYou've already chosen the best response.0
\[v \frac{ dv }{ dt } = ky ^{2} \]
 one year ago

ieatpi Group TitleBest ResponseYou've already chosen the best response.0
\[\int\limits_{}^{}Vdv = \int\limits_{}^{} ky ^{2}dy\]\[\frac{ 1 }{ 2 }V ^{2}=ky ^{1} + c\]\[V=\sqrt{2k \frac{ 1 }{ y } + c}\] So having done that, does c turn in to the initial velocity?
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
except that it is dt not dy I was thinking V= dy/dt so you could say dt= dy/V and sub in for dt in the equation.
 one year ago

ieatpi Group TitleBest ResponseYou've already chosen the best response.0
typo, should be dy
 one year ago

phi Group TitleBest ResponseYou've already chosen the best response.1
that makes it easier. to find c, plug in V0 for V and Y0 for y in your 2nd equation and solve for c
 one year ago

ieatpi Group TitleBest ResponseYou've already chosen the best response.0
ok I'll give that a shot, thanks!
 one year ago
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