Question #3fc43

2 Answers
May 19, 2016

Answer:

#t=8 s#
#AB=16 m#

Explanation:

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#"elapsed time from point O to A :" t=(OA)/v_x" "t=40/5=8" " s#

#"AB="v_y*t=2*8=16" meters"#

May 19, 2016

Answer:

The time is #t=s/(5 cos alpha)# and the distance from point #A# to point #B# is #d = (2s)/(5 cos alpha)# where #s# is the distance to shore and #alpha# is the angle which the boat points.

Explanation:

Sometimes, when we get a question like this, we don't have all of the information that is needed. In that case, we can put a place-holder (variable) in for that information and proceed with a solution. In this case, a diagram helps us decide what we have and what we need:

We are looking for the distance between points #A# and #B#, let's call that #d#. We know the speed of the boat, #5m//s#, and we know that it starts by pointing itself at point #A#. Let's call the angle that the boat is pointed in #alpha# with respect to the direction to shore, #x#.

We also know that the river is moving at #2m//s# which adds to the velocity of the boat giving the resultant velocity, #v_r#, of the boat shown in the green arrow. This resultant velocity is what causes the boat to arrive at point #B#.

Finally, we need to know the distance to the shore, let's call this #s#. The remainder is just geometric constructions.

The x-velocity of the boat is simply:

#v_x=5cos alpha#

We can calculate the time to get to shore from this and the distance to shore as:

#t=x/v_x = s/(5 cos alpha)#

The y-velocity of the boat is given by:

#v_y = 5 sin alpha -2#

From this and the time we can get the distance traveled upstream as

#c= v_y * t = (5 sin alpha - 2) s /(5 cos alpha)#

If the river wasn't moving, the boat would have reached point #A# in the same amount of time. So we can use the same approach to find the distance it would travel in this case as

#c+d = v_(yo) * t = (5 sin alpha) s /(5 cos alpha)#

We can now subtract #c# from this total to get #d#

#d = (5 sin alpha) s /(5 cos alpha)-(5 sin alpha - 2) s /(5 cos alpha)#

which simplifies to

#d = (2s)/(5 cos alpha)#