A250-m-wide river flows due east at a uniform speed of 4.3m/s. A boat with a speed of6,7m/s relative to the water leaves the south bank pointed in a direction 30° west of north.What are the (a) magnitude and (b) direction of the boat's velocity relative to the ground? (c) How long does the boat take to cross the river?

College Physics
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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@and O
are correct,
A250-m-wide river flows due east at a uniform speed of
4.3m/s. A boat with a speed of6,7m/s relative to the water leaves
the south bank pointed in a direction 30° west of north.What are
the (a) magnitude and (b) direction of the boat's velocity relative
to the ground? (c) How long does the boat take to cross the river?
IS neorrect
velacity
of the baat relatve to the ground is.
The
V br = (6:7m/s) SIN(- 30°); + (6.7 m/s) Cos (-30°)j
- 3, 35 m/s î + 5.80f
The velocity of the beat relatve to the groundd is :
Veg - V br + vr
Thus, - 3.35 m/sÎ+ 5.F0 m/s j + 4.3 m,
Piease explain in detait,
ow I am to defermne
the CORRECT directon.
Thandk qgon!
> 0.95 m/s î 5r80 im/s'j
O magnitude of Vag isi
| Vogl
(0.95)+60)
Vo.9025 + 33. 64
V34.5425
@ The time taken to
The direction of Vbg is:
Y bg X =V bgl Coso
9 = Cos/ Vbgx
Cross the wee is:
V bgy
t=250 m
5.80 m/s
Cos-!
10.95m/s)-80,65
5.88
oR 81°
43,1 S
tan Vy'
OR
tan 38080.67)
VX
0.95
Transcribed Image Text:@and O are correct, A250-m-wide river flows due east at a uniform speed of 4.3m/s. A boat with a speed of6,7m/s relative to the water leaves the south bank pointed in a direction 30° west of north.What are the (a) magnitude and (b) direction of the boat's velocity relative to the ground? (c) How long does the boat take to cross the river? IS neorrect velacity of the baat relatve to the ground is. The V br = (6:7m/s) SIN(- 30°); + (6.7 m/s) Cos (-30°)j - 3, 35 m/s î + 5.80f The velocity of the beat relatve to the groundd is : Veg - V br + vr Thus, - 3.35 m/sÎ+ 5.F0 m/s j + 4.3 m, Piease explain in detait, ow I am to defermne the CORRECT directon. Thandk qgon! > 0.95 m/s î 5r80 im/s'j O magnitude of Vag isi | Vogl (0.95)+60) Vo.9025 + 33. 64 V34.5425 @ The time taken to The direction of Vbg is: Y bg X =V bgl Coso 9 = Cos/ Vbgx Cross the wee is: V bgy t=250 m 5.80 m/s Cos-! 10.95m/s)-80,65 5.88 oR 81° 43,1 S tan Vy' OR tan 38080.67) VX 0.95
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