1) The document provides three examples of using relative vectors to solve problems involving intercepting or colliding objects. 2) The first example calculates whether two planes traveling at different velocities and positions will collide. 3) The second example calculates the direction and time for a destroyer to intercept a fishing vessel traveling at a different velocity and position. 4) The third example calculates the velocity vector and intercept time for a conservation vessel to intercept a whaling vessel traveling at a different velocity and position.

2. Example 1: At 3pm one afternoon, two planes are travelling at the same altitude. Plane A is at position 70i + 20j, travelling with a velocity (v a ) of (i - 2j) m/s and Plane B is at position 30i - 20j, travelling with a velocity (v b ) of (2i - j) m/s . Will the two planes collide if they continue on their current journeys? If we apply a vel oc ity of -v b to both planes, we effectively stop plane B and find the velocity of Plane A relative to Plane B. If this velocity takes Plane A directly towards B. (i.e. is in the same direction as ), then the planes will collide. a V b = V a - V b a V b = (i - 2j) - (2i - j) a V b = - i - j AB = b r a AB = r b - r a AB = 30i - 20j - (70i + 20j) AB = -40i - 40j AB = 40 (-i - j) or -OA + OB Thus each second, A travels (-i – j)m relative to B. After 40 seconds it will collide with B since it is travelling directly at B.

3. Example 2 : An Australian Navy Destroyer detects an Indonesion fishing vessel 412km away on a bearing of 125 °. The vessel is travelling on a bearing of 158 ° with a velocity of 16 kph. If the Destroyer travels at 24 kph, in what direction must it travel to intersect the fishing vessel and how long will it take to catch it? Lets effectively stop the Indonesian vessel and then find: n v i = v n - v i Using sine rule: sin ϕ = 0.363 ϕ = 21.29° β = 11.71° Using sin rule again | n v i |= 8.94 kph We know that | NI | = 412 km Hence t = = 46 hrs The naval ship must travel on a bearing of 146.3° (125 + 21.3°) and will take 46 hrs to intercept the fishing vessel.

4. Example 3: The Sea Shepeard (S), roaming the Antartic, position vector r s (-2i + 8j) km, picks up the whaling vessel Sushi Maru (M) on radar with a position vector (r m ) of (24i + 20j) km, and velocity vector (v m ) of (6i - 5j) km/h. If the Sea Shepeard (S) has a maximum speed of 45kph, find the velocity vector, (ai + bj), that the boat should set and the expected time of interception. We know that V s = ai + bj and that | V s | = 45 Therefore a 2 + b 2 = 45 2 ----------> Equation 1 (it is good to start with this, since we tend to forget this piece of information during the next stage)

5. If we effectively stop M by applying a negative velocity to it (i.e. find the velocity of S relative to M), we get: s V m = V s - V m s V m = (ai + bj) - (6i - 5j) s V m = (a - 6)i + (b + 5)j Finding the separation position vector. SM = m r s SM = r m – r s SM = 24i + 20j - (-2i + 8j) SM = 26i + 12j Thus for interception: t[(a - 6)i + (b + 5)j] = 26i + 12j t > 0 t(a - 6)i + t(b + 5)j = 26i + 12j t(a - 6) = 26 and t(b + 5) = 12 eliminating t gives Solving for “a” gives: 26b + 130 = 12a - 72 12a = 202 + 26b ----> Equation 2 (will check t value later)

6. Solving Simultaneously a 2 + b 2 = 45 2 and b = 12.22 or b = -25.03 since t > 0 then b = 12.22 i.e. t = 0.7 h hence a = 43.314 Thus the Sea Shepeard needs to travel with a velocity of V s = 43.14i + 12.22j and will intercept in approx 42 minutes. but are both values possible? using