The question here is how long does it take for a falling
person to reach the 90% of this terminal velocity. The computation is:
The terminal velocity vt fulfills v'=0. Therefore vt=g/c,
and so c=g/vt = 10/(100*1000/3600) = 36,000/100,000... /s. Incorporating the
differential equation shows that the time needed to reach velocity v is
t= ln [g / (g-c*v)] / c.
With v=.9 vt =.9 g/c,
t = ln [10] /c = 6.4 sec.
I hope this helps you
2y=14y-36
14y-2y=36
12y=36
y=3
Answer:
Step-by-step explanation:
W=
squaring both sides
W^2=( )^2
W^2=x-a/a
wa=x-a
a=x-a/w
The slope is change in "Y" divided by the change in "x"
1-(-3)/2-(-1)
= 4/3
Standard Form
y-y1=m(x-x1)
y-(-3)=4/3(x-(-1)
y+3=4/3x+4/3
y=4/3x-5/3
Rewrite in standard form
-4/3x+y=-5/3
Can't have fractions so multiply by 3 to get rid of the denominator
-4x+3y=5
Divide by -1 to get rid of the negative sign
4x-3y=-5
