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Date May 2010 Marks available 14 Reference code 10M.2.hl.TZ1.14
Level HL only Paper 2 Time zone TZ1
Command term Explain, Find, Hence, and Show that Question number 14 Adapted from N/A

Question

A body is moving through a liquid so that its acceleration can be expressed as

(v220032)ms2,(v220032)ms2,

where v ms1v ms1 is the velocity of the body at time t seconds.

The initial velocity of the body was known to be 40 ms140 ms1.

(a)     Show that the time taken, T seconds, for the body to slow to V ms1V ms1 is given by

T=20040V1v2+802dv.T=20040V1v2+802dv.

(b)     (i)     Explain why acceleration can be expressed as vdvdsvdvds, where s is displacement, in metres, of the body at time t seconds.

  (ii)     Hence find a similar integral to that shown in part (a) for the distance, S metres, travelled as the body slows to V ms1V ms1.

(c)     Hence, using parts (a) and (b), find the distance travelled and the time taken until the body momentarily comes to rest.

Markscheme

(a)     dvdt=v220032(=v26400200)dvdt=v220032(=v26400200)     (M1)

T0dt=V40200v2+802dvT0dt=V40200v2+802dv     M1A1A1

T=20040V1v2+802dvT=20040V1v2+802dv     AG

[4 marks]

 

(b)     (i)     a=dvdt=dvds×dsdta=dvdt=dvds×dsdt     R1

=vdvds=vdvds     AG

 

(ii)     vdvds=v2802200vdvds=v2802200     (M1)

S0ds=V40200vv2+802dvS0ds=V40200vv2+802dv     M1A1A1

S0ds=40V200vv2+802dvS0ds=40V200vv2+802dv     M1

S=20040Vvv2+802dvS=20040Vvv2+802dv     A1

[7 marks]

 

(c)     letting V = 0     (M1)

distance =200400vv2+802dv=22.3 metres=200400vv2+802dv=22.3 metres     A1

time =2004001v2+802dv=1.16 seconds=2004001v2+802dv=1.16 seconds     A1

[3 marks]

 

Total [14 marks]

Examiners report

Many students failed to understand the problem as one of solving differential equations. In addition there were many problems seen in finding the end points for the definite integrals. Part (b) (i) should have been a simple point having used the chain rule, but it seemed that many students had not seen this, even though it is clearly in the syllabus.

Syllabus sections

Topic 6 - Core: Calculus » 6.6 » Kinematic problems involving displacement ss, velocity vv and acceleration aa.
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