1. A boat needs to travel to a destination 980 km northwest of a starting point. There is an ocean current flowing towards the east at 12km/h. Assuming the maximum speed of the boat in still water is 50 km/h. What is the shortest possible time to complete the trip? حد

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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1. A boat needs to travel to a destination 980 km northwest of a starting point. There is
an ocean current flowing towards the east at 12km/h. Assuming the maximum speed of
the boat in still water is 50 km/h. What is the shortest possible time to complete the
trip?
Ad=?xmi/h
²4d=980km
Nº
Ady E-
Ś
Ade
VOE=12km/h A
VBW= SOKM/K
VBE = ?
45
VOE
540
A
VBW
At = 0.707825
Vav = 6.9 m/s
dav = ?
VBE
omn
VBW = VOE + VBE
VBG VBW VOE
VBE = √50²-12²
YBE = 48.538Km/h
48.53=980.
Ot
At=20.190h
2. A stone, thrown from ground level with a speed of 9.81 m/s at an angle of 45 above
the horizon, returns to the ground. Determine,
اد
a) the average velocity from the time of launch to the time of the maximum height.
A+= 2(9.81)Sin45
9.8
At=1.4156s (At of half)
2
4 = 0.707825
Adx = 9.812 Sinz (45)=9.8200m = 4,9/m
9.8
b) the average acceleration for the same time period in part a).
dav
6.936
Adx= 4.91m
0.707
dav = 9.800
Aav = 9.8m/s²
Vau= Ad
= 4.91
half) = 6.9 m/s
0.70782
2
Transcribed Image Text:W. + 1. A boat needs to travel to a destination 980 km northwest of a starting point. There is an ocean current flowing towards the east at 12km/h. Assuming the maximum speed of the boat in still water is 50 km/h. What is the shortest possible time to complete the trip? Ad=?xmi/h ²4d=980km Nº Ady E- Ś Ade VOE=12km/h A VBW= SOKM/K VBE = ? 45 VOE 540 A VBW At = 0.707825 Vav = 6.9 m/s dav = ? VBE omn VBW = VOE + VBE VBG VBW VOE VBE = √50²-12² YBE = 48.538Km/h 48.53=980. Ot At=20.190h 2. A stone, thrown from ground level with a speed of 9.81 m/s at an angle of 45 above the horizon, returns to the ground. Determine, اد a) the average velocity from the time of launch to the time of the maximum height. A+= 2(9.81)Sin45 9.8 At=1.4156s (At of half) 2 4 = 0.707825 Adx = 9.812 Sinz (45)=9.8200m = 4,9/m 9.8 b) the average acceleration for the same time period in part a). dav 6.936 Adx= 4.91m 0.707 dav = 9.800 Aav = 9.8m/s² Vau= Ad = 4.91 half) = 6.9 m/s 0.70782 2
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