The velocity of the current in a river is c rightarrow = 0.2 i rightarrow + 0.4
ID: 2875703 • Letter: T
Question
The velocity of the current in a river is c rightarrow = 0.2 i rightarrow + 0.4 j rightarrow km/hr. A boat moves relative to the water with velocity v rightarrow = 9 i rightarrow km/hr. What is the speed of the boat relative to the riverbed? Speed = km/hr. What angle does the velocity of the boat relative to the riverbed make with the vector v rightarrow? theta = radians What does the angle from part (b) tell us in practical terms? If you set your boat perpendicular to the y -axis at 11 km/hr, the current will take you about theta radians off course. If you set your boat parallel to the y - axis 9 km/hr, the current will take you about theta radians off course. If you set your boat parallel to the x - axis at 9 km/hr, the current will take you about theta radians off course. The angle does not tell us anything in practical terms. If you set your boat perpendicular to the x - axis at 11 km/hr, the current will take you about theta radians off course.Explanation / Answer
a>
The velocity of the boat relative to the riverbed is
u = v + c
u = 9i + (0.2i + 0.4j)
u = 9.2i + 0.4j
Hence the magnitude of the vector u is :
u^2 = 9.2^2 + .4^2 = 84.8
The speed of the boat relative to the riverbed is :
s = ||u|| = sqrt(u^2)
s = sqrt(84.8) = 9.21 km/h
b>
Let 't' be the angle between u and v
u•v = ||u||*||v||*cost
cost = (u•v)/(||u||*||v||)
where
u•v =(9.2i + 0.4j)•9i = 82.8
||u||*||v|| = 9.21*9 = 82.89
Hence
cost = 82.8/82.89 = 0.9989
=> t = cos^-1(.9989)
=> t = .047 radians
4.3º
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