Physics Question

[TheJourney]

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A boat moves with a speed of 20 m/s in still water. It is placed on the shore of a rive 346m wide. The river flows at a rate of 10m/s. In what direction must the boat be pointed to reach a point on the opposite shore that is directly across from the point of its departure in a time of 20 seconds?

A) The boat must be pointed downstream at an angle of 30 degrees from the direction perpendicular to the bank.
B) The boast must be pointed upstream at an angle of 45 degrees from the direction perpendicular to the bank.
C) The boat must be pointed upstream at an angle of 60 degrees from the direction perpendicular to the bank.
D) The boat must be pointed upstream at an angle of 30 degrees form the direction perpendicular to the bank.
E) the boat must be pointed downstream at an angle of 60 degrees from the direction perpendicular to the bank.

Highlight the following text for the answer choice. The answer is choice D

Can anyone explain to me how they personally arrived at the answer.

 

[syoung]

Hopefully you can infer that you should not be pointing your boat downstream, as you would never reach the opposite bank directly across from the point of departure

Draw a triangle in a river for the velocities.

The distance covered from bank to bank is 346 m. That must be the relative horizontal distance traveled by the boat, and is related to its relative horizontal velocity.

The triangle is a right angled one with the hypotenuse being the speed of the boat in still water (20 m/s) and the vertical component being the velocity of the river (10 m/s)

Solving for the last component of horizontal velocity results in 17.3 m/s which fits the 346 m in 20 s horizontally.

Solve then for the angle, 30 degrees upstream.

 

[labqi]

Sorry. Why do you use the horizontal velocity. Aren't you moving upstream? Could you by any chance post up a drawing?

 

[Chrisz]

Hi, brother, this question actually does not need any calculation, because it gives too much info. it says the flow rate is 10m/s, so your boat should be positioned such that the horizontal velocity counters the 10m/s flow rate. The horizontal velocity is 10m/s as well. (horizontal is the axis which river flows). Since you also know the boat goes 20m/s in still water, this is hypotenuse of the vertical and horizontal velocity. So sin(theta)=10/20, so theta=30 degrees. No need to calculate the vertical velocity, which is 346/20. So answer D

 

[NextStepTutor_2]

Hi labqi and TheJourney, Great questions. I've got my response, here http://forums.studentdoctor.net/threads/specific-science-questions.1053558/

Hope this helps!

 

[TheJourney]

Thank you so much for clarifying that!