What differentiates an athletic trainer from a clinical exercise physiologist?
2 answers:
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Answer:
Attached below are the sketches
answer :
c) G(s) = 100 / ( s + 100 )
d) y'(t) + 100Y(s) = 100 X(s)
e) g(t) = e^-100t u(t)
Explanation:
a) Sketch the bode plot
The filter here is a low pass filter
b) Sketch the s-plane
attached below. pole ( s ) is at 100
c) write the transfer function of the filter
Transfer function ; G(s) = 100 / ( s + 100 )
d) write the differential equation
Y(s) / X(s) = 100 / s + 100
Y(s) [ s + 100 ] = 100 X(s)
= sY(s) + 100Y = 100 X(s)
∴ differential equation = y'(t) + 100Y(s) = 100 X(s)
e) write out the unforced transient response
g(t) = e^-100t u(t)
f) write out the frequency response
attached below
Answer
given,
Speed of vehicle = 65 mi/hr
= 65 x 1.4667 = 95.33 ft/s
e = 0.07 ft/ft
f is the lateral friction, f = 0.11
central angle,Δ = 38°
The PI station is
PI = 250 + 50
= 25050 ft
using super elevation formula
r = 1568 ft
As the road is two lane with width 12 ft
R = 1568 + 12/2
R = 1574 ft
Length of the curve
L = 1044 ft
Tangent of the curve calculation
T = 542 ft
The station PC and PT are
PC = PI - T
PC = 25050 - 542
= 24508 ft
= 245 + 8 ft
PT = PC + L
= 24508 + 1044
=25552
= 255 + 52 ft
the middle ordinate calculation
MO = 85.75 ft
degree of the curvature
D = 3.64°