question 2: relative velocity - web viewuse pythagoras’ theorem to get the length of this...

Question 2: Relative Velocity

Category 1: Two Independent Bodies(i) General Questions

Higher Level

2012 (a)

1

Category 1: Two Independent Bodies(ii) Interception

If both bodies have an i and a j component then for interception to occur you must make use of the fact that either the i component or the j component will be the same.

2010 (a)

2

Category 1: Two Independent Bodies:(iii) Shortest Distance

Usually at higher level one of the cars will have to be brought to the junction – so which one?Technically it doesn’t matter, but the solution is more straightforward if (when one car is at the junction) the other is also on the x–axis.2009 (a)

3

2011 (a)

4

2006 (b)

5

2004 (b)

6

Category 1: Two Independent Bodies –

(iv) Distance from Intersection (Higher Level)2008 (a)

7

2009 (a)

8

2005 (b)

9

2003 (b)

10

2004 (a)

11

2004 (b)

12

Two Independent Bodies – (v) Distance from each other

To solve this one draw a circle around the ‘stationary’ object whose radius corresponds to the distance given in the question (say 10 km). Now draw the Vab vector.Where this cuts the circle corresponds to A first coming within 10 km of B and where it cuts the circle on the way out corresponds to A being more than 10 km of B.Use Pythagoras’ Theorem to get the length of this chord. Now knowing this length and the velocity Vab, work out the time.

2012 (b)

13

2007 (a)

14

Body and Carrier exam questions

2004 (a)

15

2011 (b)

2007 (b)

16

Body and Carrier General2006 (a)

17

2005 (a)

18

2009 (b) {very difficult}

19

Exam Questions

2003 (a)

20

2010 (b)

21

2008 (b)

22