1. ## relative velocity angle

question 1.

part i distance traveled by man d=s*t .6*20=12 distance traveled by woman =.4*20=8 i cant do part ii. ive tried setting up a triangle with there distances i just worked out and you are given the angle tan^-1 3/4 and i have tried the sine rule to work it out but the right answer is 60 degrees and i am not getting it. previous similar worked examples were solved using the sine rule

2. ## Re: relative velocity angle

The distances you have calculated are correct but they are not the sides of a triangle.

Redraw your diagram, so that the woman ends up due south of the man. Your diagram will consist of 2 triangles. Then try again.

Come back if you get stuck again.

4. ## Re: relative velocity angle

Do you know if they are the exact sizes of the sides triangles. 10/sinx = 8/sin(tan^-1 3/4). Does not work

5. ## Re: relative velocity angle

No it doesn't work because the triangle(s) you need to work with does not include 8 and 10 as sides.

Angle ACB = arctan(3/4)

CA = 10 (distance apart at start)

Using basic trig (or recognising a 3:4:5 = 6:8:10 right angled triangle). AB=6 and BC=8

CP = 12 (distance swum by man) and AR = 8 (distance swum by woman)

Now concentrate on triangles ABQ and RPQ (which are similar).

See if that helps.

6. ## Re: relative velocity angle

man swimmer starts at (0,0) and swims due W at 0.6 m/s ...

$x_1(t)=-0.6t$

$y_1(t)=0$

woman swimmer starts at (-8,6) and swims at an angle $\alpha$ S of W at 0.4 m/s ...

$x_2(t) = -8 -0.4\cos{\alpha} \cdot t$

$y_2(t) = 6 - 0.4\sin{\alpha} \cdot t$

at $t=20$ sec, the woman swimmer is due S of the man swimmer ...

$x_2(20)=x_1(20)$

$-8-8\cos{\alpha}=-12 \implies \cos{\alpha}=\dfrac{1}{2} \implies \alpha = 60^\circ$

for part (iii) ...

$d=\sqrt{[x_2(t)-x_1(t)]^2 + [y_2(t)-y_1(t)]^2}$

expand and simplify the resulting time quadratic under the radical and find the time for minimum distance by either calculus or non-calculus means.

7. ## Re: relative velocity angle

Ok thanks. But is there not any way to work it out with the triangle's

8. ## Re: relative velocity angle

There is no way to tell distance AQ or PQ. All i know is the size of AB