W.+1. A boat needs to travel to a destination 980 km northwest of a starting point. There isan ocean current flowing towards the east at 12km/h. Assuming the maximum speed ofthe boat in still water is 50 km/h. What is the shortest possible time to complete thetrip?Ad=?xmi/h²4d=980kmNºAdy E-ŚAdeVOE=12km/h AVBW= SOKM/KVBE = ?45VOE540AVBWAt = 0.707825Vav = 6.9 m/sdav = ?VBEomnVBW = VOE + VBEVBG VBW VOEVBE = √50²-12²YBE = 48.538Km/h48.53=980.OtAt=20.190h2. A stone, thrown from ground level with a speed of 9.81 m/s at an angle of 45 abovethe horizon, returns to the ground. Determine,ادa) the average velocity from the time of launch to the time of the maximum height.A+= 2(9.81)Sin459.8At=1.4156s (At of half)24 = 0.707825Adx = 9.812 Sinz (45)=9.8200m = 4,9/m9.8b) the average acceleration for the same time period in part a).dav6.936Adx= 4.91m0.707dav = 9.800Aav = 9.8m/s²Vau= Ad= 4.91half) = 6.9 m/s0.707822

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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? حد

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

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Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

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Conservation of Energy

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Question

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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