a. X = 3m/s.
Y = 4m/s.
tanA = Y/X = 4/3 = 1.3333.
A = 53.1o,CCW
A = 90-53.1 = 36.9o East of North.
Vb = 3/cos53.1 = 5m/s @ 53.1o =
Magnitude and direction.
b. d = 128/sin53.1 = 160.1 m.
t = d/V = 160m / 5m/s = 32 s.
c. The wind causes the boat to head 36.9o East of North. To compensate for the wind, the boat must head 36.9o West of North.
A ferryboat, whose speed in still water is 4.00m/s, must cross a river whose current is 3.00m/s. The river runs from west to east and is 128m wide. The boat is pointed north.
a) If the boat does not compensate for the flow of the river water and allows itself to be pushed off course, what would be the new velocity of the boat? State both the magnitude and the direction of the velocity of the boat relative to the riverbank.
b) In part (a), what is the velocity of the boat across the river? How long would it take the boat to travel across the river?
c) The boat now compensates for the river current so that it travels straight across the river to the north without being pushed off course.
2. Relevant equations
let Vbw represent velocity of boat relative to the water.
let Vbs represent velocity of the boat relative to the shore.
let Vws represent velocity of the water relative to the shore.
Vbs=Vbw+Vws
pythagorean theorem using the variables Vbs²=Vbw²+Vws²
V=d/t (not sure how to write the vector symbols for velocity and distance, but I know they're supposed to be there and that is what V and d mean)
3. The attempt at a solution
a) Vbs²=Vbw²+Vws²
Vbs²= (4.00m/s)²+(3.00m/s)²
Vbs²=16.00m/s+9.00m/s
Vbs²=25.00m/s
Vbs=*√25.00m/s
Vbs= 5.00m/s {41º E of N}
Θ = tan-¹ (3.00m/s / 4.00m/s)
= 41º
∴ If the boat does not compensate for the flow of the river water and allows itself to be pushed off course, the new velocity of the boat would be 5.00m/s {41º E of N}
b)
In part (a) the velocity of the boat across the river is 4.00m/s
V=d/t
t=d/V
t= 128m/4.00m/s
t= 32s
∴ it would take the boat 32s to travel across the river.
c) Vbs=Vbw+Vws
Vbs=4.00m/s {at some angle} + 3.00m/s {E}
to find the angle use inverse Sine function.
θ= Sine-¹(3.00m/s / 4.00m/s)
θ= 54°
∴ If the boat now compensates for the river current so that it travels straight across the river to the north without being pushed off course, it should be headed at 54° angle.
Please correct any of my work if I have made any mistakes at any point in solving this problem. Thanks for your help.
1 answer
1 answer