Physics - Relative Motion Homework Help

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SUMMARY

The discussion focuses on solving a relative motion problem involving a ferryboat crossing a river. The ferryboat has a speed of 4.00 m/s in still water and encounters a current of 3.00 m/s. When not compensating for the current, the boat's resultant velocity is calculated to be 5.00 m/s at an angle of 41º east of north. The time taken to cross the 128m wide river is determined to be 32 seconds. When compensating for the current, the boat must head at an angle of 54° to travel straight across.

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  • Familiarity with the Pythagorean theorem
  • Knowledge of trigonometric functions (sine, cosine, tangent)
  • Ability to solve basic motion equations (V=d/t)
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Homework Statement



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.

Homework 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)

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.
 
Physics news on Phys.org
Your methods are all correct, but in each case the angle seems a bit too large. atan(0.75) = 36.9 degrees and asin(.75) = 48.6 degrees. That last must be about right because sin(45 deg) = sqrt(1/2) = .707...
 

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