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Unformatted text preview: 2.3 ‘2, 3,. 23 23.(a) Measure the positive direction of motion downward. Based on Newton’s
second law, the equation of motion is given by dv _ {—0.75v+mg, 0 <1: < 10 m.— _
dt —1212+mg, t> 10 Note that gravity acts in the positive direction, and the drag force is resistive.
During the ﬁrst ten seconds of fall, the initial value problem is dv/dt = —v / 7.5 + 32,
with initial velocity 11(0) : 0 ft/s. This differential equation is separable and linear,
with solution 005) 2 240(1 — (ft/7'5). Hence 11(10): 176.7 ft/s. (b) Integrating the velocity, with 330%) = 0, the distance fallen is given by
$05) = 2401: + 1800 64/”  1800. Hence 2:00) 2 1074.5 ft. (c) For computational purposes, reset time to t = 0. For the, remainder of the
motion, the initial value problem is dv/dt = —32'u/ 15 + 32, with speciﬁed initial
velocity 71(0) = 176.7 ft/s. The solution is given by 71(t’)~.: 15 + 161.7(3—32 V15. As
75 —> oo, 110$) —> ’UL : l5 ft/s. (d) Integrating the velocity, with 56(0) 2 1074.5, the distance fallen after the parachute
is open is given by 33(2‘.) 2 15t — 75.86—32’5/15 + 1150.3. To ﬁnd the duration of the second part of the motion, estimate the root of the transcendental equation
15T _ 75867327715 + 1150.3 2 5000. The result is T = 256.6 s. (6) Velocity "ae’elscity
180 ,2»:
150 MEI
120' l I l. ma 1,
3 a “ , a 'w
513 mu 11.50 200 2513 5 if? 1.. n 29.(a) Let 2: represent the height above the earth’s surface. The equation of motion
is given by 772% = —G&% , in which G is the universal gravitational constant. The symbols M and R are the mass and radius of the earth, respectively. By the
chain rule, ~ 2710121 — —G———Mm
d3: .— (R + x)2 ' This equation is separable, with min = —G’M (R + :13) ‘2dx. Integrating both sides,
and invoking the initial condition 12(0) 2 V2912, the solution is v2 = 2GM(R + :c)‘1 + 29R — ZGM/R.
From elementary physics, it follows that g = GM / R2. Therefore 71(95): @(R/VR + 90).
(Note that g = 78, 545 Dill/1112.) (b) We now consider dac/dt = #29 (R/VR +32). This equation is also separable,
With \/ R + a: d 2 M29 Rdt. By deﬁnition of the variable :17, the initial condition is
93(0) = 0. Integrating both sides, we obtain use) = (3th + git/2))?“ — R. Setting the distance a:(T) + R = 240, 000, and solving for T, the duration of such
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 TERRELL,R

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