# Thread: relative velocity and interception

1. ## relative velocity and interception

a patrol boat, travelling at 40km/h on a brearing of 030 degree, sees a suspect craft travelling at 15 km/h due north. It intercepts the craft one hour later. What was the position of the craft relative to the patrol boat when it was first sighted?

2. Originally Posted by yanmingfu
a patrol boat, travelling at 40km/h on a brearing of 030 degree, sees a suspect craft travelling at 15 km/h due north. It intercepts the craft one hour later. What was the position of the craft relative to the patrol boat when it was first sighted?
note the sketch. let the patrol craft start at the origin at t = 0.

patrol boat's position after 1 hour ...

$x = 40\cos(60)$

$y = 40\sin(60)$

suspect craft's position one hour earlier ...

$x = 40\cos(60)$

$y = 40\sin(60) - 15$

find the bearing/range to the suspect craft's initial position.

3. the answer is 28km at 45.5 degree but I don't know how it comes

4. Originally Posted by yanmingfu
the answer is 28km at 45.5 degree but I don't know how it comes
hi

Vp= velocity of patrol boat , Vc=velocity of craft

cVp=velocity of c relative to P

cVp=Vc-Vp

=15 j -20 i -20\sqrt{3} j

=-20 i +(15-20\sqrt{3})j

s=vt , since t=1 , s=v

cRp= displacement of craft relative to patrol boat

= |cVp |

$=\sqrt{(-20)^2+(15-20\sqrt{3})^2}$

$\approx 28.03 km$

$\tan \theta=\frac{15-20\sqrt{3}}{-20}$

$\theta=44.5^o$

the direction given in the answer is measured from south .

5. Originally Posted by yanmingfu
the answer is 28km at 45.5 degree but I don't know how it comes
what do you know about solving this problem?

6. There are a number of different ways to do this problem. Since you made no attempt to solve it yourself, we don't know which would be appropriate for you. If you are given a problem like this, you must have had some instruction in them. Do you know how to "solve triangles", using the sine and cosine laws? Do you know how to separate a vector into its components?