Solving Distance and Direction-Based Homework Problems

[munkachunka]

[/B]

Homework Statement

rower a wishes to cross a river directly to the oposite bank (east) a distance of 30mtrs (90 degree's to the shore) the current is flowing NS at 2.6m/s and he can sustain a rowing speed of 3m/s

1. calculate the direction with respect to the current flow he needs to row (degrees) in order to achieve his goal.

2. Rower B is heading south at a relative speed of 3m/s with respect to the current and his position is 80m NNE relative to rower A's start point. determine the closest point of the two rowers.

Homework Equations

d=s/t
trig

The Attempt at a Solution

1. time to cross 30 meters = 30m/3ms = 10 seconds.
distanced pushed to the south = 2.6m/s*10 = 26mtrs

using trig i find the angle to be 40.91.

taking north as the reference point, the angle rower a needs to go is 90-40.91 = 49.08degrees.

2. I am unsure how to find the closest point but did the following.

made the total speed equal to 3m/s+2.6m/s = 5.6m/s
The angle from rower a to rower b's start position is 22.5deg.
the NS distance between the rowers is 85.44m.

I am guessing that it has something to do with Vectors but I am unsure, could someone please check what I have done and then give me a pointer as to the correct way to approach part 2.

thankyou

 

[learningphysics]

Part 1 is wrong. To get directly to the east, his net velocity in the north-south direction must be zero... You need to be able to add his velocity vector which has a magnitude of 2.6... which the river's velocity vector 3... and the result should be directly eastward... use vector geometry to calculate the direction his velocity vector needs to be pointed...

For part 2, I don't understand what 80m NNE means. Did they give 22.5 or did you calculate that?

 

[munkachunka]

I am not to sure where I am going here, I don't understand what you have said

I think the velocity vector for rower a is 39.7m, 40.91degrees,(using pythagoras) I am stuck as to what to do next.
I thought I would just add this angle up stream which would then negate the rower being pushed down stream.

for part B I was given 22.5deg, the NNE is taken from rower a's start position which works out to be 22.5

 

[learningphysics]

I am not to sure where I am going here, I don't understand what you have said

I think the velocity vector for rower a is 39.7m, 40.91degrees,(using pythagoras) I am stuck as to what to do next.
I thought I would just add this angle up stream which would then negate the rower being pushed down stream.

Suppose he goes at 40.91 degrees... so his vertical velocity is 3sin40.91 - 2.6 = -0.635m/s... so by the time he crosses, he will be displaced downstream... the net vertical velocity should be zero.

for part B I was given 22.5deg, the NNE is taken from rower a's start position which works out to be 22.5

So is the initial horizontal distance between A and B 80cos67.5, and the vertical distance is 80sin67.5 ?

 

[munkachunka]

thanks for your help,

I understand what you are saying about the rower being displaced using the angle I gave, I have worked it out roughly by entering different angles into my calculator, I think the answer is around 60degrees,

I got that by trial and error, I am not getting the method behind this, I know it must be simple because there isn't much info given at the start of the question but I can't see the way to go about it.

stu

 

[learningphysics]

Yup, that's the right answer. The trick is to draw a triangle representing the vectors... it is a right triangle.

The rower's velocity vector added to the south bound vector of the river... results in a vector to the east...

That means the rower's velocity vector is the hypoteneuse... the river's velocity is one side... and the resultant is the third side...

From that triangle you can find the angle of the hypoteneuse from the horizontal...

I'm certain your textbook will have example problems for these type of problems...

 

[munkachunka]

thanks

When I do the triangle I get the angle to be 40.91Deg

I use tan^-1*2.6/3. I do this using basic trig but I guess is not the way to do it.

using pythagoras to get the velocity SQRT(3^2 +2.6^2) = 3.96

I can't think of another way to do this although I know it is wrong because of the way we proved it yesterday, i have been looking at the following website which has a similar example half way down the page.
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l1f.html

thankyou

 

[learningphysics]

3 is the hypoteneuse of the triangle. so the net velocity is [tex]\sqrt{3^2 - 2.6^2}[/tex]

The direction is [tex]sin^{-1}(\frac{2.6}{3})[/tex]

 

[learningphysics]

You can also think about it with displacement vectors... Suppose the the rower is traveling at an angle theta...

Suppose the travel time is t... the river pushes him downstream by 2.6t...

So the rower's displacement without the river needs to have a vertical component of 2.6t upstream.

ie the component of the rower's velocity upstream is 2.6. You know the net velocity is of the rower is 3 (hypoteneuse of the triangle)... the vertical component is 2.6...

 

[learningphysics]

You know the vertical component of the rower's needs to be 2.6m/s (to cancel the river)... You also know that the net velocity of the rower is 3m/s... From these two you get get the direction of the rower's velocity...

So at this point, we don't need to worry about the river... just analyze the rower's velocity vector... the horizontal component + vertical component = the net velocity...

So the triangle just analyzes the rower alone without the river...

 

[learningphysics]

You can also just look at it as the vector sum of the rower's velocity... and the river's velocity...

draw the rower's velocity vector at an arbitrary angle... add to that vector, a downwards vector (the river)... the resultant needs to be directly to the right (otherwise the rower will end up downstream)... do you see how the right triangle forms with the rower's velocity as the hypoteneuse?

 

[learningphysics]

seems like they want the angle from the river's velocity... so that's 90 + 60.07 = 150.07 counterclockwise from the river's velocity...

 

[munkachunka]

oh right,

I thought I had to take the 3m/s going directly east and then the 2.6m/s south, I didnt realize that the 3m/s is the hypotenuse.

I understand now that when the rower is sustaining his 3m/s he is being pushed south and hence it is the hypotenuse.

I can see this now but would probably get a similar question wrong in the future

I take it the approach on this website is wrong!
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l1f.html

I now know the angle is 60 degrees, do I express this as 30 deg because it want the answer with respect to the current flow? I took north as 0 deg, I worked out the magnitude of the hypotenuse to be 60 meters i.e 30m/1.49m/s = 20 secs ... 20*3m/s = 60m

now I have the answer for the first part I am after trying to sort out part 2,

I have worked out the horizontal distance of b to be 73.91m and the vertical to be 30.61m I am not sure how this help though.
do I simply add the two vectors together B=80 angle 67.5 and A=60 angle 60?

 

[munkachunka]

I have had a more in depth look at the website mentioned above, I can see the subtle differencesin the questions now and can see that it is correct.

stu

 

[munkachunka]

I can also see what you mean about the angle being 150degrees, I guess 0 degrees is south because that is the direction of the river.

thanks

 

[learningphysics]

I can also see what you mean about the angle being 150degrees, I guess 0 degrees is south because that is the direction of the river.

thanks

No prob.

 

[munkachunka]

now I have the answer for the first part I am after trying to sort out part 2,

I have worked out the horizontal distance of b to be 73.91m and the vertical to be 30.61m I am not sure how this help though.
do I simply add the two vectors together B=80 angle 67.5 and A=60 angle 60?

 

[learningphysics]

now I have the answer for the first part I am after trying to sort out part 2,

I have worked out the horizontal distance of b to be 73.91m and the vertical to be 30.61m I am not sure how this help though.
do I simply add the two vectors together B=80 angle 67.5 and A=60 angle 60?

Try to get the coordinates of A (xa,ya), and the coordinates of B (xb,yb)... they will depend on time...

I'd use the start point of A as (0,0)... and the start point of B is (73.91,30.61)... so that is at t=0...

Can you get the coordinates at an arbitrary time? Use the velocity of A and B...

From the coordinates, you can get the distance at an arbtrary time... then you need to find the t when this is a minimum...

 

[munkachunka]

I'd use the start point of A as (0,0)... and the start point of B is (73.91,30.61)... so that is at t=0...

using t = 20 i get

A = X30.61,Y-0
B = X30.61,Y-38.09 (assuming total velocity is 3m/s + 2.6m/s)


I am unsure what you mean by this statement "Can you get the coordinates at an arbitrary time?"
I am unsure

 

[learningphysics]

I hope I didn't confuse you with my use of "net velocity"...

I should really distinguish as velocity with respect to the river... and velocity with respect to the ground...

What is the velocity of A with respect to the ground?

You can write the position of A as:

(x,y) = (0,0) + (vx,vy)t = (vx*t,vy*t)

where (vx,vy) is the velocity of A with respect to the ground.

 

[munkachunka]

I am really sorry, I am just not getting this,
using t = 20

a =(vx*t,vy*t) (1.49*20,2.6*20) = 29.8,52

b =(vx*t,vy*t) (0*20,5.6*20) = 0,112 (I don't think this is correct, the horizontal velocity is 0 because rower B is heading directly south, however rower B is not a 0 on the grid.

stu

 

[learningphysics]

I am really sorry, I am just not getting this,
using t = 20

That's ok.

a =(vx*t,vy*t) (1.49*20,2.6*20) = 29.8,52

Hope I didn't confuse about the velocity of the rower...

The velocity of the rower with respect to the ground is (1.497, 0) (this is what someone standing on the ground looking at the rower will see)

The velocity of the rower with respect to the river... ie the velocity with which he'd be going if the river was still... is 3m/s in the direction we got in part a... 3cos60.07 in the x direction. 3sin60.07 in the y direction... (1.497, 2.6)

Before we go on, let me know if this part makes sense... I know I used the words "net velocity" before in two contradictory ways... so I won't say use those words again. It's better to use velocity wrt river... and velocity wrt ground to distinguish the two...

For part 2, what we need is the velocity with respect to ground... Let me know if velocity wrt ground vs. velocity wrt river makes sense...

 

[munkachunka]

thankyou for the explanation, that makes total sense now you put it that way.

so for B is it (0, 5.6) wrt ground i.e 0 in the x direction and 3m/s+2.6m/s for the Y??

stu

 

[learningphysics]

thankyou for the explanation, that makes total sense now you put it that way.

so for B is it (0, 5.6) wrt ground i.e 0 in the x direction and 3m/s+2.6m/s for the Y??

stu

Cool! yeah, exactly... but I'd use (0, -5.6)

taking north as positive, and east as positive...

So for part 2, you need the position at any t (any time between 0 and 20s)...

So for A:

(xA(t), yA(t)) = (0,0) + (1.49,0)t

This equation is just another way of saying:
final position = initial position + velocity*time

the only difference is that we're using two-dimensional vectors...

If this notation is confusing deal with x and y separately... so
xA(t) = 0 + 1.49t = 1.49t
yA(t) = 0

then in coordinate form (xA(t), yA(t)) = (1.49t, 0)

Here t is not referring to the 20s... but to any time between 0 and 20s... For example, at 5s, A is at (1.49*5, 0) = (7.45,0)

Let me know if this part makes sense. we can plug in t into (1.49t, 0) to get the coordinates of A at any time between 0 and 20s.

 

[munkachunka]

yeah, that makes total sense, I understand that the time can plug into the formula to give the position at any given time,

i.e for B if the time was 5 sec.. then the co-ordinates are 0, -28

without trial and error I am still unsure how to get the closest point together

thanks again

 

[learningphysics]

That's cool... we're almost there... To minimize, we'll need to get the distance as a function of time... then either use derivatives or the minimum of a parabola... to get the minimum distance... have you studied calculus?

So the position for A is (1.49t, 0)...

BTW I'm getting the initial position for B as (30.61, 73.9)... do you agree? just want to make sure we're on the same page, because it seems earlier in the thread we switched the x and y coordinates...

So at any time t... for B the position is (30.61, 73.9 - 5.6t)... do you see how I get this?

 

[munkachunka]

yeah I see where that has come from.

A= (1.49t, 0)... I understand that as the value of t increases so does the x component

B= (30.61, 73.9 - 5.6t)... I understand that as the value of t increases the y value reduces

I have studied a bit of calculus but not a great deal, hit me with it if you please but go gentle

 

[learningphysics]

Cool! Now, you've got 2 points A and B...

A= (1.49t, 0)
B= (30.61, 73.9 - 5.6t)

What is the distance between these two points? Just plug the expressions into the distance formula...

 

[munkachunka]

I have the formula...

sqrt((x2-x1)^2+(y2-y1)^2) == using A= (1.49t, 0) and B= (30.61, 73.9 - 5.6t)

sqrt(0-30.61)^2+(0-73.9)^2 =79.98 where t=0
sqrt(1.49-30.61)^2+(0-73.9-5.6)^2 = 74.24 where t=1

I can see the distance coming down, I can see we are nearly there now

 

[learningphysics]

I have the formula...

sqrt((x2-x1)^2+(y2-y1)^2) == using A= (1.49t, 0) and B= (30.61, 73.9 - 5.6t)

sqrt(0-30.61)^2+(0-73.9)^2 =79.98 where t=0
sqrt(1.49-30.61)^2+(0-73.9-5.6)^2 = 74.24 where t=1

I can see the distance coming down, I can see we are nearly there now

Cool... but don't plug in numbers for t... you need the algebraic expression for distance with t in there...

 

[munkachunka]

Ok

like ...sqrt(1.49t-30.61)^2+(0-73.9-5.6t)^2

 

[learningphysics]

Ok

like ...sqrt(1.49t-30.61)^2+(0-73.9-5.6t)^2

exactly... be careful of the second parentheses... you made a mistake with signs...

that's the formula for distance... we need to find the time when this distance is minimum...

The square root makes this a little messy... so what we can do is find the time when distance^2 is minimum... that's the same as when distance is minimized... this just allows us to get rid of the square root...

Multiply out and simplify the expression for distance^2...

 

[munkachunka]

d^2=1.49^2*t^2-30.61^2+-73.9^2-5.6^2*t^2 =

d^2=2.22*t^2-936.9+-5461.21-31.36*t^2 =

d^2 = -6427.25*t^4

not sure I have done that right beacuse the answer should be positive,

I can't see where the sign is wrong

 

[learningphysics]

distance^2 = (1.49t-30.61)^2+(0-(73.9-5.6t))^2

distance^2 = (1.49t-30.61)^2+(0-73.9+5.6t)^2

distance^2 = (1.49t-30.61)^2 + (-73.9 + 5.6t)^2

distance^2 = (1.49t-30.61)^2 + (5.6t -73.9)^2

You didn't multiply it out correctly... remember:

[tex](a+b)^2 = a^2 + 2ab + b^2[/tex]
[tex](a-b)^2 = a^2 - 2ab + b^2[/tex]

 

[munkachunka]

ah yes, my silly mistake, it is getting late here now

distance^2 = (1.49t-30.61)^2 + (5.6t -73.9)^2

==d^2 = 1.49t^2-2*1.49t*30.61+-30.61^2 + 5.6t^2-2*5.6t*-73.9+-73.9^2

= d^2 = 1.49t^2-91.2t+936.9 +5.6t^2+827.7t+5461.21

=d^2 = 7.09t^2+736.5t+6398.11

is this right so far?

I have to get some sleep now, it is 2300 this side of the pond, I look forward to finishing this tomorrow with a fresh mind!

thankyou so much for your help today, it has been great