1. ## relative velocity interception

a sailing boat is travelling at a constant speed of 4 km hr^-1 in a direction 35 degrees north of east. a motorboat wishes to intercept the sailing boat is travelling at a speed of 10 km hr^-1. at a certain point the motorboat is 2km northwest of the sailing boat. in what direction should the motor boat head in order to intercept the sailing boat

what ive got so far: velocity of sailing boat=3.27i+2.3j velocity of motor boat=(10cosx)i+(10sinx)j if i get there relative velocity the j components should cancel out so 2.3=10sinx x=13.297 degrees sadly this is not the right answer. the right answer is 21.8 degrees south of east.

2. ## Re: relative velocity interception

Let the motorboat be at the origin at $t=0$ with the saiboat 2km southeast of the motorboat at position $r_s=2\cos(-45)i + 2\sin(45)j$.

Sailboat velocity ... $v_s=4\cos(35)i + 4\sin(35)j$.

Motorboat velocity ... $v_m=10\cos{\theta} i + 10\sin{\theta} j$

Let $t$ be the time of interception.

In the E/W direction ...

$10\cos{\theta} \cdot t = 2\cos(-45) + 4\cos(35) \cdot t \implies t = \dfrac{\cos(-45)}{5\cos{\theta}-2\cos(35)}$

In the N/S direction ...

$10\sin{\theta} \cdot t = 2\sin(-45) + 4\sin(35) \cdot t \implies t = \dfrac{\sin(-45)}{5\sin{\theta}-2\sin(35)}$

Since $\sin(-45) = -\cos(-45)$ ...

$5\cos{\theta} - 2\cos(35) = 2\sin(35) - 5\sin{\theta}$

$\cos{\theta}+\sin{\theta} = \dfrac{2}{5}\bigg[\cos(35)+\sin(35) \bigg]$

Let the right side = constant, $k$, and square both sides ...

$\sin(2\theta) + 1 = k^2$

$\theta = \dfrac{1}{2} \arcsin(k^2-1) \approx -21.8^\circ$ or $21.8^\circ$ S of E.

3. ## Re: relative velocity interception

im also trying to solve in a different way cause sometimes you are not given the distance between the ships. here is a example of the way im trying to do it from a different question. for the way they are working it out could you use the same way for the question i posted.?
im trying to work it out this way but it wont work

4. ## Re: relative velocity interception

the text re-orients the i and j directions as shown in the diagram ...

5. ## Re: relative velocity interception

Thanks that's what I was looking for. I kept getting my answer 23.2 why are you taking 45 degree away? Is it to get it back to normal I and j components

6. ## Re: relative velocity interception

Originally Posted by markosheehan
Thanks that's what I was looking for. I kept getting my answer 23.2 why are you taking 45 degree away? Is it to get it back to normal I and j components
Reorientation of the i direction from its normal direction East to a new direction Southeast is equivalent to a $-45^\circ$ rotation ... so, the new direction for $\vec{i}$ is $-45^\circ$. Since $\theta$ is relative to that new direction, the direction of $\theta$ relative to due East is actually $-45^\circ + \theta$.

Since $\theta > 0$, I just subtracted $45^\circ$