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Unformatted text preview: Problem 13.6 The position of a point during the inter
val oftime from I = 0tot = 6 s isgiven by s = 7%? +
632 + 4: m. (a) What is the maximum velocity during this interval
of time, and at what time does it occur?
(b) What is the acceleration when the velocity is a maximum?
. du .
Solution: Maximum velocily occurs where a : E : 0 (it could be a minimum)
. . V da ,
.r : filll + 6:2 + 4’ m Thls occurs at! = 4 5. At this pomt E — —3 so we have a maxmium.
v : ~§r2 + 123+ 4 “hrs (:1) Max velocity is at r :4 s. where u : 28 m/s and (b) a z 0 mls?‘
a : —3: + 12111152 Problem 13.15 The acceleration of a point is a = 60: — 36:2 ms? When I = 0, 3 2 0 and v 2 20 ft/s.
What are position and velocity as a function of time? Solution: The velocity is
v = fad! +C. = fuse: — 36:2)+C1= 30;2 —1213 + 6.. At t : 0, v : 20 W3, hence C1 = 20, and the velocity as a function
of time is u = 30:2 —12:3 + 20 fI/s. The position is
32fum+cz :[oott 12:3+20) +c2 :10:3 — 3:4 + 20: + C2‘ At r : O, s : 0. hence C2 = 0, and the position is s = I0r3 — 3:4 + 20r n Problem 13.28 Determine the distance traveled during
its takeoff by the airplane in Problem I327. Solution: for 0 g r _<_ 5 s an}: {.2 2
a =( S ).v+(3 fu'sz). v: (6 Us )%+{3 fu'sz): 53 53 M5 5) : 30 fu's, 3(5 3) : 62.5 ft
for 5 s 51 5 30 s a = 9 ms? 0 : {9 M2)“ , 5 s) + 30 flis. 7 _5 
.a‘ z {9 fus2)% + (30 mm: , 5 s} + 62.5 r‘: => 5(30 5) = 3625 fl Problem 13.48 In Problem 13.47, what distance does
the airliner require to take off? Solution: 0!
a : vi?” : (14 W52} — (0.0003 ft_l)u2
X 200 11/5 m1” .1
f —2—_12 2] d”
u (14 ftfs ) — (0.0003 ft )v u ...
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 Spring '08
 N/A
 Dynamics

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