Give a formula for the velocity as a function of the displacement given that when s = 1 metre, v = 2 m s 1. (7)

1. The acceleration of a body is given in terms of the displacement smetres as a= 1 2 2  s s .

(a) Give a formula for the velocity as a function of the displacement given that when s= 1 metre, v= 2 m s–1.

(7)

(b) Hence find the velocity when the body has travelled 5 metres.

(2) (Total 9 marks)

2. Solve the differential equation 2 d d

y x y

x  = 1, given that y= 0 when x= 2. Give your answer in the form y= f(x).

(Total 6 marks)

3. (a) Show that the solution of the differential equation

x y d d = cos xcos2y, given that y= 4 π

when x= π, is y= arctan (1 + sin x).

(5)

(b) Determine the value of the constant afor which the following limit exists

2 2 π 2 π ) sin 1 arctan( lim           x a x x

and evaluate that limit.

(12) (Total 17 marks)

4. The population of mosquitoes in a specific area around a lake is controlled by pesticide. The rate of decrease of the number of mosquitoes is proportional to the number of mosquitoes at any time t. Given that the population decreases from 500 000 to 400 000 in a five year period, find the time it takes in years for the population of mosquitoes to decrease by half.

5. A particle moves in a straight line in a positive direction from a fixed point O. The velocity vm s−1, at time tseconds, where t0, satisfies the differential equation





. 50 1 d d v v2 t v   

The particle starts from O with an initial velocity of 10 m s−1.

(a) (i) Express as a definite integral, the time taken for the particle’s velocity to decrease from 10 m s−1to 5 m s−1.

(ii) Hencecalculate the time taken for the particle’s velocity to decrease from 10 m s−1 to 5 m s−1.

(5)

(b) (i) Show that, when v0, the motion of this particle can also be described by the differential equation





50 1 d d v2 x v  

 where xmetres is the displacement from O. (ii) Given that v= 10 when x= 0, solve the differential equation expressing xin terms

of v.

(iii) Henceshow that v= . 50 tan 10 1 50 tan 10 x x   (14) (Total 19 marks)

6. A certain population can be modelled by the differential equation t y d d

= k ycos kt, where yis the population at time thours and kis a positive constant.

(a) Given that y= y₀when t= 0, express yin terms of k, tand y₀.

(5)

(b) Find the ratio of the minimum size of the population to the maximum size of the population.

(2) (Total 7 marks) 7. The acceleration in m s–2of a particle moving in a straight line at time tseconds, t≥ 0, is given by the formula a= v

2 1

8. (a) Solve the differential equation x y x y y _d d e e cos e 2

 = 0, given that y= 0 when x= π.

(7)

(b) Find the value of ywhen x= 2 π .

(1) (Total 8 marks)

9. Find yin terms of x, given that (1 + x3)

2 π and tan 2 d d 2   x y y x y when x= 0. (Total 7 marks) 10. A body is moving through a liquid so that its acceleration can be expressed as

          32 200 2 v m s–2,

where vm s–1is the velocity of the body at time tseconds. The initial velocity of the body was known to be 40 m s–1.

(a) Show that the time taken, Tseconds, for the body to slow to Vm s–1is given by T=



 40 2 2 80 1 200 V _v dv. (4)

(b) (i) Explain why acceleration can be expressed as s v v d d , where sis displacement, in metres, of the body at time tseconds.

(ii) Hence find a similar integral to that shown in part (a) for the distance, Smetres, travelled as the body slows to Vm s–1.

(7)

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

(3) (Total 14 marks)

11. A skydiver jumps from a stationary balloon at a height of 2000 m above the ground. Her velocity, vm s–1, tseconds after jumping, is given by v= 50(1 – e–0.2t).

(a) Find her acceleration 10 seconds after jumping.

(3)

(b) How far above the ground is she 10 seconds after jumping?

(3) (Total 6 marks) 12. A jet plane travels horizontally along a straight path for one minute, starting at time t= 0, where

tis measured in seconds. The acceleration, a, measured in m s–2, of the jet plane is given by the straight line graph below.

(a) Find an expression for the acceleration of the jet plane during this time, in terms of t.

(1)

(b) Given that when t= 0 the jet plane is travelling at 125 m s–1, find its maximum velocity in m s–1during the minute that follows.

(4)

(c) Given that the jet plane breaks the sound barrier at 295 m s–1, find out for how long the jet plane is travelling greater than this speed.

(3) (Total 8 marks)

1. (a) a= 1 2 2  s s a= s v v d d M1 1 2 d d 2   s s s v v s s s v v d 1 2 d 2





 _ M1 2 2 v  = ln│s2+ 1│ + k A1A1

Note:Do not penalize if kis missing. When s= 1, v= 2  2 = ln 2 + k M1  k= 2 – ln 2 A1                 2 2 1 ln 2 ln 2 1 ln 2 2 2 2 s s v A1 (b) EITHER 2 2 26 ln 2 2   v M1  v2= 2 ln│13│+ 4 4 13 ln 2   v A1 OR 2 2 v = ln│26│+ 2 – ln 2 M1 v2= 2 ln│26│+ 4 – 2 ln 2 v= 2ln2642ln2 A1 [9]

2. 1 d d , 1 d d 2 2      y x y x y x y x Separating variables (M1) x x y y d 1 d 2   A1 arctan y= ln x+ c A1A1 y= 0, x= 2  arctan 0 = ln 2 + c –ln 2 = c (A1) arctan y= ln x –ln 2 = ln 2 x y=       2 ln tan x A1 N3 [6]

3. (a) this separable equation has general solution

∫sec2y dy= ∫cos x dx (M1)(A1)

tan y= sin x+ c A1

the condition gives tan

4 π

= sin π + c  c= 1 M1

the solution is tan y= 1 + sin x A1

(b) the limit cannot exist unless a= arctan        2 π sin 1 = arctan 2 R1A1

in that case the limit can be evaluated using l’Hopital’s rule (twice) limit is                     2 π 2 lim 2 π 2 ) ) sin 1 (arctan( lim 2 π 2 π x y x x x x M1A1

where yis the solution of the differential equation

the numerator has zero limit (from the factor cos xin the differential equation) R1 so required limit is 2 lim 2 π y x    M1A1 finally,

y″ = –sin xcos2y– 2 cos xcos ysin y× y′(x) M1A1 since 5 1 2 π cos       y A1 y″ = 2 π at 5 1   x A1

the required limit is 10

1

 A1

4. Let the number of mosquitoes be y. ky t y   d d M1





y k t yd d 1 M1 lny= –kt + c A1 y= e–kt+c y= Ae–kt when t= 0, y= 500 000  A= 500 000 A1 y= 500 000e–kt when t= 5, y= 400 000 400 000 = 500 000e–5k M1 5 4 = e–kt –5k= 5 4 ln k= 5 4 ln 5 1  (= 0.0446) A1 250 000 = 500 000e–kt M1 2 1 = e–kt kt   2 1 ln t= 2 1 ln 5 4 ln 5 = 15.5 years A1 [8]

5. (a) (i) EITHER

Attempting to separate the variables (M1)





50 d 1 d 2 t v v v    (A1) OR Inverting to obtain

v

t

d

d

(M1)



2



1 50 d d v v v t    (A1) THEN









           





10 5 2 5 10 2 ₁ d 1 50 d 1 1 50 v v v v v v t A1 N3 (ii)









        sec 101 104 ln 25 sec 732 . 0 t A2 N2 (b) (i) x v v t v d d d d  (M1)

Must see division by v(v> 0) A1





50 1 d d v2 x v    AG N0

(ii) Either attempting to separate variables or inverting to obtain v x d d (M1)







v x v d 50 1 1 d 2 (or equivalent) A1

Attempting to integrate both sides M1

arctan v=  x C

50 A1A1

Note: Award A1 for a correct LHS and A1 for a correct RHS that must include C.

When x= 0, v= 10 and so C= arctan10 M1

(iii) Attempting to make arctan vthe subject. M1 arctan v= arctan10  50 x A1 v= tan        50 10 arctan x M1A1

Using tan (AB) formula to obtain the desired form. M1

50 tan 10 1 50 tan 10 x x v    AG N0 [19] 6. (a) t y d d = kycos (kt) y y d = k cos(kt)dt (M1)





 k kt t y y d ) cos( d M1 lny= sin(kt) + c A1 y= Aesin(kt) t= 0  y0= A (M1)  y= y0esin kt A1 (b) – l ≤ sin kt≤ 1 (M1) y0e–1≤y≤y0e1 so the ratio is e 1 : e or 1 : e2 A1 [7]

7. v t v 2 1 d d   A1 t v v d 2 1 d





  (A1) ln v=  tc 2 1 (A1) v=             t c t A 2 1 2 1 e e (A1) t= 0, v= 40, so A= 40 M1 v= 2t 1 e 40  (or equivalent) A1 [6] 8. (a) rearrange x x y x y x y y y y _d 0toobtain cos d e e d d e e cos e 2 e 2    (M1) as 2 sin(2 ) ₁ 4 1 2 1 d 2 ) 2 cos( 1 d cos x x



 x x x x C



M1A1 and



eyeeydyeey C₂ A1

Note:The above two integrations are independent and should not be penalized for missing Cs.

a general solution of x x C x y x y y y      e e 2 e ) 2 sin( 4 1 2 1 is 0 d d e e cos A1

given that y= 0 when x= π, C= e

2 π e ) π 2 sin( 4 1 2 π _e0     (or – 1.15)(M1)

so, the required solution is defined by the equation

                    2 π e ) 2 sin( 4 1 2 1 ln ln e 2 π e ) 2 sin( 4 1 2 1 e x x y x x y or A1 N0 (or equivalent) (b) for x= _              4 π e ln ln , 2 π y (or –0.417) A1 [8]

9. (1 + x3) x x x y y y x y y d 1 2 tan d tan 2 d d 3 2 2





 _   M1 x x x y y y d 1 3 3 2 d sin cos 3 2





 _ (A1)(A1) C x y  ln1 3  3 2 sin ln A1A1

Notes:Do not penalize omission of modulus signs. Do not penalize omission of constant at this stage.

EITHER 0 1 ln 3 2 2 π sin ln  CC M1 OR |sin y| = 3 2 3 1 x A  , A= eC 1 0 1 2 π sin 3 2 3     A A M1 THEN y= arcsin          3 2 3 ) 1 ( x A1

Note: Award M0A0 if constant omitted earlier.

[7] 10. (a) _             200 6400 32 200 d d v2 v2 t v (M1) v v t V T d 80 200 d 40 2 2 0





  _ M1A1A1 T= 200 v v V ₈₀ d 1 40 2 2



_ AG (b) (i) a= t s s v t v d d d d d d   R1 = s v v d d AG

(ii) 200 80 d d  2  2  v s v v (M1) v v v s V S d 80 200 d 40 2 2 0





  _ M1A1A1 v v v s V S d 80 200 d 40 _{2 2} 0





 _ M1 S= v v v V ₈₀ d 200



40 _{2 2}  A1 (c) letting V= 0 (M1) distance = v v v d 80 200 40 0 2 2



_ = 22.3 metres A1 time = v v 80 d 1 200 40 0 2 2



_ = 1.16 seconds A1 [14]

11. (a) a= 10e–0.2t (M1)(A1)

at t= 10, a= 1.35 (m s–2) (accept 10e–2) A1 (b) METHOD 1 d=



10   0 2 . 0 d ) e 1 ( 50 t t (M1) = 283.83... A1

so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1

METHOD 2

s = ∫50(1 – e–0.2t)dt= 50t+ 250e–0.2t(+ c) M1

Taking s= 0 when t= 0 gives c= –250 M1

So when t= 10, s= 283.3...

so distance above ground = 1720 (m) (3 s.f.) (accept 1716 (m)) A1

[6]

12. (a) equation of line in graph a= t 60 25  + 15 A1          15 12 5 t a

(b) 15 12 5 d d    t t v (M1) v= 2 24 5 t  + 15t+ c (A1) when t= 0, v= 125 m s–1 v= 2 24 5 t  + 15t+ 125 A1

from graph or by finding time when a= 0

maximum = 395 m s–1 A1

(c) EITHER

graph drawn and intersection with v= 295 m s–1 (M1)(A1)

t= 57.91 – 14.09 = 43.8 A1 OR 295 = 2 24 5 t  +15t+ 125  t= 57.91...; 14.09... (M1)(A1) t= 57.91… – 14.09… = 43.8 (8 30 ) A1 [8]