RELATIVE VELOCITY INDEPENDENT BODIES (INTERCEPTION) Applied Maths

Transcription

1 RELATIVE VELOCITY INDEPENDENT BODIES (INTERCEPTION) Applied Maths

2 PREMIUM VERSION PREVIEW

3 2013 LCHL Question 2 (b) An aircraft P, flying at 600 km h 1, sets out to intercept a second aircraft Q, which is a distance away in a direction west 30 south, and flying due east at 600 km h 1. Find the direction in which P should fly in order to intercept Q. Depending on your initial approach this is unbelievably easy OR quite tricky! The easy version here. Mathematically fine so presume this was accepted for the 25 marks!! Diagram Interception occurs when the direction of the relative velocity vector, Ԧv ab is equal to the direction of the relative position vector, Ԧr ba. Ԧv PQ α Ԧv P 600 Clearly an Isosceles Triangle and therefore α = 30 and the angle P needs to fly is West 60 South! 600 Ԧv Q

4 2013 LCHL Question 2 (b) An aircraft P, flying at 600 km h 1, sets out to intercept a second aircraft Q, which is a distance away in a direction west 30 south, and flying due east at 600 km h 1. Find the direction in which P should fly in order to intercept Q. Resolve the velocity vectors into their Ԧi and Ԧj components Ԧv P = 600 cos α Ԧi 600 sin α Ԧj Ԧv Q = 600 Ԧi Bit more work in this version. Ԧv PQ = Ԧv P Ԧv Q Ԧv PQ = 600 cos α Ԧi 600 sin α Ԧj 600 Ԧi Ԧv PQ = 600 cos α Ԧi 600 sin α Ԧj For interception to occur the direction of v PQ is West 30 South. Direction of Particle Ԧj component tan θ = Ԧi component 600 sin α tan 30 = 600 cos α = 600 sin α 600 cos α sin α = 13 cos α + 1 cos α + 1 = 3 sin α cos α = 3 sin α 2 cos 2 α + 2 cos α + 1 = 3sin 2 α cos 2 α + 2 cos α + 1 = 3 1 cos 2 α cos 2 α + 2 cos α + 1 = 3 3cos 2 α 4cos 2 α + 2 cos α 2 = 0 2cos 2 α + cos α 1 = 0 2 cos α 1 cos α + 1 = 0 2 cos α 1 = 0 cos α + 1 = 0 2 cos α = 1 cos α = 1 2 α = 60 The angle P needs to fly is West 60 South! cos α = 1 α = 180 Not possible in this situation cos 2 α + sin 2 α = 1

5 2014 LCHL Question 2 (a) Two ships X and Z are observed from a coastguard station. With distances measured in kilometres and speeds in kilometres per hour, they have the following displacement and uniform velocity vectors: Ship X Y Z Time 14:00 14:30 15:00 Displacement 2i + 7j 6i + 9j 12i + 9j Velocity 4i + 5j 3i + 4j 2i + 6j Marking scheme method much quicker on the next slide! This probably more useful though for the majority of questions. Just 20 Marks for this (a) part. (i) Prove that if X and Z continue with their uniform velocities they will collide. Find the time of the collision. Interception occurs when the direction of the relative velocity vector, Ԧv ab is equal to the direction of the relative position vector, Ԧr ba. More specifically: Ԧv ab = k Ԧr ba We must find the position of X at 15:00 Ԧr X = 2Ԧi + 7Ԧj + 4Ԧi + 5Ԧj Ԧr X = 6Ԧi + 12Ԧj Position of Z relative to X at 15:00 Ԧr ZX = Ԧr Z Ԧr X Magnitude xԧi + yԧj = x 2 + y 2 Time = Distance Speed Ԧv XZ = 2Ԧi Ԧj Ԧv XZ = Ԧv XZ = 5 Time = 45 5 Time = 3 hours Ԧv XZ = Ԧv X Ԧv Z Ԧv XZ = Ԧv X Ԧv Z = 4Ԧi + 5Ԧj 2Ԧi + 6Ԧj = 2Ԧi Ԧj Ԧr ZX = Ԧr Z Ԧr X = 12i + 9j 6Ԧi + 12Ԧj = 6Ԧi 3Ԧj = 3 2Ԧi Ԧj 3 Ԧv XZ = Ԧr ZX Ԧr ZX = 6Ԧi + 3Ԧj Ԧr ZX = Ԧv ZX = 45 N.B. 3 hours after 15:00 Collision occurs at 18:00 Therefore the ships collide.

6 2014 LCHL Question 2 (a) Two ships X and Z are observed from a coastguard station. With distances measured in kilometres and speeds in kilometres per hour, they have the following displacement and uniform velocity vectors: Ship X Y Z Time 14:00 14:30 15:00 Displacement 2i + 7j 6i + 9j 12i + 9j Velocity 4i + 5j 3i + 4j 2i + 6j (i) Prove that if X and Z continue with their uniform velocities they will collide. Find the time of the collision. Alternate Method Much quicker!! If they are to collide then their displacements must be equal at some future time, t hours after 14:00 and t 1 hours after 15:00. Ԧr X = Ԧr Z 2Ԧi + 7Ԧj + 4Ԧi + 5Ԧj t = 12Ԧi + 9Ԧj + 2Ԧi + 6Ԧj t 1 2Ԧi + 7Ԧj + 4Ԧit + 5Ԧjt = 12Ԧi + 9Ԧj + 2Ԧit + 6Ԧjt 2Ԧi 6Ԧj 2 + 4t = t 2 2t = 8 t = 4 hours Particles will collide at 18:00

7 2014 LCHL Question 2 (a) Ship X Y Z Time 14:00 14:30 15:00 Displacement 2i + 7j 6i + 9j 12i + 9j Velocity 4i + 5j 3i + 4j 2i + 6j (ii) At the instant of the collision, ship Y changes course and then proceeds directly to the scene of the collision at its original speed. Find the time, to the nearest minute, at which Y will arrive at the scene of the collision. We must find the position of Y at 18:00, 3.5 hours after Y s position was 6Ԧi + 9Ԧj. Ԧr Y = 6Ԧi + 9Ԧj Ԧi + 4Ԧj Ԧr Y = 6Ԧi + 9Ԧj Ԧi + 14Ԧj Ԧr Y = 16.5Ԧi + 23Ԧj We must find the position of X at 18:00, 4 hours after X s position was 2Ԧi + 7Ԧj. Ԧr X = 2Ԧi + 7Ԧj + 4 4Ԧi + 5Ԧj Ԧr X = 2Ԧi + 7Ԧj + 16Ԧi + 20Ԧj Ԧr X = 18Ԧi + 27Ԧj Position of X relative to Y at 18:00 Ԧr XY = Ԧr X Ԧr Y Ԧr XY = Ԧr X Ԧr Y = 18Ԧi + 27Ԧj 16.5Ԧi + 23Ԧj = 1.5Ԧi + 4Ԧj Magnitude xԧi + yԧj = x 2 + y 2 Ԧr XY = Ԧr XY = 73 2 m This is the distance between the scene of the collision and Y at 18:00 Calculate Y s original speed. Ԧv Y = 3i + 4j Ԧv Y = = 5 km/h Time = Distance Speed 73 2 = hours 5 = 51 minutes Y will arrive at the scene at 18: 51

8 PREMIUM VERSION PREVIEW