- #1
apchemstudent
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However, I am stuck on this problem
Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>=0. After her parachute opens, her velocity satisfies the differential equation dv/dt = -2v -32, with initial condition v(0) = -50.
A) Find an expression for v in terms of t, where t is measured in seconds.
well I have problems right from the start...
dv/(-2v-32) = dt
integrate both sides --> ln(-2v-32) = -2t
I solve for v and i got v = e^(-2t)/-2 - 16 + C
C = -33.5
However, i could also have done it like this
dv/(-2(v+16)) = dt
bring the -2 over and integrate it to get --> ln(v+16) = -2t
solve for v = e^(-2t) -16 + C
C = -35
Why are they different?! i must be doing something wrong...
b) Terminal velocity is defined as lim t->infinity v(t). Find the terminal velocity of the skydiver to the nearest foot per second.
well for the first equation i got
i get -49.5 feet per second, since the 1/infinity = 0
while the second one is -51 feet per second... which is most likely not correct.
c) It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
well for both cases i got t = negative seconds. I really don't think this is right. Can some one please help me figure out my mistake? I am really lost thanks...
Let v(t) be the velocity, in feet per second, of a skydiver at time t seconds, t>=0. After her parachute opens, her velocity satisfies the differential equation dv/dt = -2v -32, with initial condition v(0) = -50.
A) Find an expression for v in terms of t, where t is measured in seconds.
well I have problems right from the start...
dv/(-2v-32) = dt
integrate both sides --> ln(-2v-32) = -2t
I solve for v and i got v = e^(-2t)/-2 - 16 + C
C = -33.5
However, i could also have done it like this
dv/(-2(v+16)) = dt
bring the -2 over and integrate it to get --> ln(v+16) = -2t
solve for v = e^(-2t) -16 + C
C = -35
Why are they different?! i must be doing something wrong...
b) Terminal velocity is defined as lim t->infinity v(t). Find the terminal velocity of the skydiver to the nearest foot per second.
well for the first equation i got
i get -49.5 feet per second, since the 1/infinity = 0
while the second one is -51 feet per second... which is most likely not correct.
c) It is safe to land when her speed is 20 feet per second. At what time t does she reach this speed?
well for both cases i got t = negative seconds. I really don't think this is right. Can some one please help me figure out my mistake? I am really lost thanks...