Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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SECTION 13.4 MOTION IN SPACE: VELOCin' AND ACCELERATION D 171
80 sin 9 .--- 5 4.9 (-- 5 ) = 15 => 100 tan () - 7.65625 sec 2 fJ = 15. Replacing sec 2 9 with tan 2 8 + 1 gives
4cos9 4 cosfJ
7.65625 tan 2 9 - 100 tanfJ + 22.65625 = 0. Using the quadratic formula, we have tan 9 ~ 0.230635, 12.8306 =>
B ~ 13.0°, 85.5°. So for 13.0° < 9 < 85.5°, the rock will land beyond the near city wall. The base of the far wall is
located at (600, 0) which the rock hits if (80 cos fJ)t = 600 => t = 2
15 n and (80sin 9)t - 4.9t
2 = 0 =>
COSu
. 15 ( 15 ) 2
80sm0 · --n - 4.9 - (} = 0 => 600tan9- 275.625sec 2 fJ = 0 =>
2
2cOS!7
COS
275.625 tan 2 9 - 600tan0 + 275.625 = 0. Solutions are tan(}~ 0.658678, 1.51819 => (} ~ 33.4°, 56.6°.· Thus the
rock lands beyond the enclosed city ground for 33.4° < (} < 56.6°, and the angles that allow the rock to land on city ground
are 13.0° < () < 33.4~, 56.6° < 9 < 85.5°. If you consider t~at the rock can hit the far wall and bounce back into the city, we
calculate the angles that cause the rock to hit the top of the wall at (600, 15): (80 cos 9)t = 600 => t = ~ and
2 cos(}
(80sinfJ)t- 4.9t 2 = 15 => 600tan0 -275.625sec 2 (} = 15 => 275.625 tan 2 (} - 600 tan(}+ 290.625 = o.
Solutions are tan 9 :::::: 0. 727506, 1.44936 =>
13.0° < (} < 36.0°' 55.4° < (} < 85.5° .
(} :::::: 36.0°, 55.4°, so the catapult should be set with angle fJ where
31. Here a (t) = - 4 j - 32 k so v (t) = -4tj - 32t k + v o = - 4tj- 32tk + 50 i +80 k = 50i- 4tj + (80 - 32t) k and
r (t) = 50t i- 2t 2 j + (BOt- 16e) k (note that r o = 0). The ball lands when the z-component ofr (t) is zero
and t > 0: SOt - 16t 2 = 16t{5 - t) = 0 => t = 5. The position of the ball then is
r (5) = 50(5) i - 2{5) 2 j + [80(5) - 16(5) 2 ] k = 250 i - 50 j or equivalently the point (250, - 50, 0). This is a distance of
J250 2 + ( -50)2 + 0 2 = )65,000 :::::: 255ft from the origin at an angle oftan- 1 (~) :::::: 11.3° from the eastern direction
toward the south. The speed of the ball is lv (5)l = I 50 i- 20j - 80 k l = J50 2 + ( - 20)2 + ( -80)2 = v'9300:::::: 96.4 ftls.
33. (a) After t seconds, the boat will be 5t meters west of point A. The ve lo~ i ty
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of the water at that location is 4 ~ 0 (5t)(40- 5t)j . The velocity of the
boat in still water is. 5 i, so the resultant velocity of the boat is
v(t) = 5 i + 4 ~ 0 (5t)( 40- 5t)j = 5i + ( ~t- fG-t2 ) j . Integrating, we obtain
r (t) = 5t i + (~tl - -fGt 3 ) j +C. lfwe place the origin at A (and consider j
to coincide with the northern direction) then r (O) = 0 => C = 0 and we have r (t) = 5t i + (~t 2 -
reaches the east bank after 8 s, and it is located at r(8) = 5(8)i + ( ~(8) 2 -
downstream.
1
e) j . The boat
16 -fG (8?) j = 40 i + 16j. Thus the boat is 16 m
(b) Let Q be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by
5(cos Q) i + 5(sin Q) j . At t seconds, the boat is 5(cos Q)t meters from the west bank, at which point the velocity
of the water is . 1 ~ 0 [5(cos Q)t][40- 5(cos Q)t] j . The resultant velocity of the boat is given by
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