Slide 1 : RELATIVE VELOCITY Consider 2 cars A & B passing a point P on a straight road. A is travelling at constant velocity of 30m/h and B is travelling at constant velocity of 10m/h. A B 30m/h 10m/h Question: From point of view of person sitting in car B what velocity does A appear to be travelling at?
Our intuition tells us 20m/h.
This can be expressed mathematically as the velocity of A relative to or * NOTE: This also applies for cars travelling in opposite directions. Say A is travelling east at 30m/h and B is coming against it in opposite direction at 10m/h.
Slide 2 : A B 30m/h 10m/h Again our intuition tells us that to an observer at B, A appears to be travelling towards him at a velocity of 40m/h. Using the equation we get (Minus sign indicates opposite direction) When velocities are not along same line and relative velocity needs to be calculated the equation can still be applied if velocities are resolved in directions E.G. 2 ships leave from same point – one ship is travelling at 10m/h in direction SW, the other is travelling at 20m/h in direction SE. N E W S
Slide 3 : can also be shown graphically using laws of vector addition (triangle or parallelogram law)
Slide 4 : Mag and Direction of Velocity Say (velocity of A relative to B) works out to be Q) What is magnitude and direction? 3 4 N S E W ANS Mag 5m/s
Dir E 53.130N NOTE Speed = Dist/Time
Rel Speed = Rel Dist/Time
Slide 5 : EXAM QUESTION TYPES
Relative Velocity – Type 1 – ‘Plane in Still Air’ Example Airport X is 400km due north of airport Y. A wind blows from the west at 60km/h and an aircraft can maintain a speed of 500km/h in still air. In what direction must the aircraft fly and how long does the journey take from Y to X? 400km X Y Basic Ideas
The plane is ‘headed’ in a certain direction travelling at 500km/h
This direction is such that the wind ‘blows’ the aeroplane into the north/south direction, that is, from Y to X Resultant Velocity = (500km/h in ‘headed’ direction) + (60km/h East)
(North Direction) Using trigonometry of right angled triangle 500 60 NOTE: This type
was asked as half-question in
2006, 2004, 2001, 1999, 1992
All that is required is addition of vectors and basic trigonometry of right angled triangle
Slide 6 : Relative Velocity – Type 2 – Crossing a River A boat can travel at 13m/s in still water. It has to travel across a stream which is 78m wideand flowing at where is the unit vector along the straight banks. How long will the crossing take
1) By the quickest route?
2) By the shortest route? Solution:
So what’s the difference?
(i) If the boat puts all it’s force into going across it will cross in the shortest time
In this case 78m 5 12 13 Time = Dist/Speed
= 78/12 = 6.5secs Asked 2007,2005, 2006,1998, 1995, 1993 as half-question
Slide 7 : Relative Velocity – Type 3 – ‘Wind appears to blow’ * This is a very popular question and has been asked in 2008,2003, 1997, 1993, 1991, 1989 Example 2003 2(a)
A woman travelling north at 10km/h finds that the wind appears to blow from the west. When the woman trebles her speed, the wind appears to blow from the north-west.
Find the velocity of the wind.
Method Treat 2 parts of journey separately. Identify the different vectors for each part and write them in terms of . Note that actual velocity of wind is constant in both parts of journey. N E S W Part 1 x=a y=10
Slide 8 : Part 2 (y-30) x Note: Since this negative, magnitude is -(y-30)
Slide 9 : Relative Velocity – Type 4 – Shortest Distance e.g. 2002 Q2(a) Two boats, B & C, are each moving with constant velocity. At a certain instant boat B is 10km due west of boat C. The speed and direction of boat B relative to boat C is 2.5m/s in direction 600 south of east. Calculate the shortest distance between the boats to the nearest metre.
Solution: The shortest distance is worked out from trigonometry using principle that if the relative velocity vector is plotted then the shortest distance is the perpendicular distance from the vector to the ‘stationary’ object
i.e. in this example dist is from C to 10km 2.5m/s d B C 600 Came up 2008, 2005, 2004, 2003, 2002, 2000, 1996
Slide 10 : Relative Velocity – Type 5 – Angle of Closest Approach e.g. 1999 Q2(b) Two ships A & B move with constant speeds of 48km/h and 60km/h respectively. At a certain instant ship B is 30km west of A and is travelling due south. Find the direction ship A should steer in order to get as close as possible to ship B.
In this type of question you are not given one of the angles so you call it (say) . Then you work out the angle ( ) of the relative velocity vector in terms of and optimize this by differentiation and putting = 0. 48km/h 60km/h 30km/h A B
Slide 11 : B
Slide 12 : Relative Velocity – Type 6 – Interception/Collision Course e.g. 1998 Q2 a The driver of a speedboat travelling in a straight line at 20m/s wishes to intercept a yacht travelling at 5m/s in a direction 400 east of north. Initially the speedboat is positioned 5km south-east of the yacht. Find the direction of the speedboat if it intercepts the yacht.
In this type of question the key is to realise that the relative velocity vector must be along the displacement vector in order for collision to occur. A B In the diagram if the position of A & B relative to A is taken as usual to mean that A appears stationary then it is obvious that VAB must be along the line AB in direction of A for collision to occur.
i.e. A B
Slide 13 : 1998 Q2(a) 450 500 400 VSY y SE S W 450 VS(20m/s) Vy(5m/s) has to be along displacement line shown for collision to occur By sine rule