Hello steve989 Originally Posted by

**steve989** Hi, I have a mechanics question here I can't quite get.

A destroyer sights a ship at a point with position vector 600(3**i **+ **j**)m relative to it and moving with velocity 5**j** m/s. The destroyer alters course so that it moves with speed **v** m/s in the direction of the vector 4**i** + 3**j**. Find **v** so that the destroyer intercepts the ship and the time to the interception.

Any help would be greatly appreciated.

Thanks

Denote the velocity of the destroyer by $\displaystyle \vec{v_D}$ and the velocity of the ship by $\displaystyle \vec{v_S}$. Then:$\displaystyle \vec{v_D}=\frac{v}{5}\Big(4\vec i + 3 \vec j\Big)$, since this vector has magnitude $\displaystyle v$.

and$\displaystyle \vec{v_S} = 5\vec j$

So the velocity of the ship relative to the destroyer is$\displaystyle \vec{v_S}-\vec{v_D} = -\frac{4v}{5}\vec i +\Big(5-\frac{3v}{5}\Big)\vec j$

Therefore after time $\displaystyle t$, the position of the ship relative to the destroyer is:$\displaystyle 600(3\vec i + \vec j) -\frac{4vt}{5}\vec i +\Big(5-\frac{3v}{5}\Big)t\vec j$

$\displaystyle =0\vec i + 0\vec j$ when the destroyer intercepts the ship

Solving for $\displaystyle v$ and $\displaystyle t$ gives:

$\displaystyle t = 390$ and $\displaystyle v = \frac{75}{13}$

But check my working!

Grandad