2012 Extension 2 Resisted Motion
1 like400 views
1) This document is a 2012 HSC exam solution for a question about an object sinking in liquid. 2) The equation of motion for the object is derived using partial fractions, showing the velocity v over time t. 3) The displacement x over time is then derived by substituting the expression for v into the equation relating v and x. 4) The specific question is answered by substituting t = 4 seconds into the expressions for v and x to find how far the object has sunk after 4 seconds.
2012 Extension 2 Resisted Motion
- 1. Extension II Mathematics: Mechanics 2012 HSC Resisted Motion Brett M. Bujeya John Paul College June 23, 2013 This work is licensed under the Creative Commons 3.0 License. Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 1 / 7
- 2. 2012 HSC, Question 13Question 13 (15 marks) Use a SEPARATE writing booklet. (a) An object on the surface of a liquid is released at time t = 0 and immediately sinks. Let x be its displacement in metres in a downward direction from the surface at time t seconds. The equation of motion is given by 2 dv v = 10 , dt 40 where v is the velocity of the object. (i) (ii) (iii) t 20 e 1( )Show that v = .t e + 1 dv dv 400 Use = v to show that x = 20logedt dx 400 v How far does the object sink in the first 4 seconds? . 4 2 2 2 Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 2 / 7
- 3. 2012 HSC, Question 13(i) Solution dv dt = 10 v2 40 dv dt = 400 v2 40 40 v 0 dv 400 v2 = t 0 dt Using partial fractions, 40 dv 400 v2 = 40 dv (20 + v)(20 v) = 1 20 + v + 1 20 v dv t 0 dt = v 0 1 20 + v + 1 20 v dv t = ln 20 + v 20 v v 0 t = ln 20 + v 20 v et = 20 + v 20 v 20(et 1) = v(1 + et ) v = 20(et 1) 1 + et Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 3 / 7
- 4. 2012 HSC, Question 13(ii) Solution v dv dx = 10 v2 40 v dv dx = 400 v2 40 x 0 dx = 40 v 0 v 400 v2 dv x = 20 ln(400 v2 ) v 0 x = 20 ln 400 v2 400 x = 20 ln 400 400 v2 Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 4 / 7
- 5. 2012 HSC, Question 13(iii) Solution When t = 0, v = 0 and x = 0 When t = 4, v = 20(e4 1) e4 + 1 and x = 20 ln 錚 錚 400 400 400(e41)2 (e4+1)2 錚 錚 = 20 ln e4 + 1 2 4e4 Hence, the particle moves 20 ln e4 + 1 2 4e4 metres in the 鍖rst 4 seconds. Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 5 / 7
- 6. Notes From the Marking Centre I i. In better responses, after 鍖nding the expression dt dv = 40 400 v2 , candidates used partial fractions and correct integration to obtain t = ln 20 + v 20 v . In better responses, candidates demonstrated their competency by changing the subject to obtain the required expression for v. In weaker responses, some candidates attempted to use the result for dx a2 x2 without resorting to partial fractions. In many of these responses, candidates did not obtain the correct result of this integration. ii. Having obtained dx dv = 40v 400 v2 , the majority of candidates handled this part well and obtained the required result. Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 6 / 7
- 7. Notes From the Marking Centre II iii. This part was again well done. The most successful approach was to substitute t = 4 into v = 20(et 1) et + 1 and then substitute this expression into x = 20 ln 400 400 v2 . Another approach was to integrate using dx dt = 20(et 1) et + 1 . Only a small number of candidates used this approach, and of them very few successfully obtained the required result. Brett M. Bujeya 2012 HSC - Resisted Motion June 23, 2013 7 / 7