Differential equations — Paper 3 · A Level Mathematics - Further 9231

7 A particle P moving in a straight line has displacement x m from a fixed point O on the line at time t s. -2 200 100 The acceleration of P, in ms , is given by 2 - 3 for x 2 0 . When t = 0 , x = 1 and P has velocity x x -1 10 ms directed towards O. -1 10 ( 1 - 2 x) (a) Show that the velocity v ms of P is given by v = . [5] x … … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that x and t are related by the equation e -40 t = ( 2x - 1) e 2 x -2 and deduce what happens to x as t becomes large. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) 3 2 d 100 200 d v v x x x = − + 2 2 50 200 2 v A x x = − + M1 A1 Correct equation and attempt to integrate Correct 1, 10: 200 x v A = = − = M1 Use initial condition ( ) 2 2 2 100 2 1 x v x − = M1 Rearrange to find 2v ( ) 10 2 1 x v x − = ± and take negative sign to meet initial condition, so ( ) 10 1 2x v x − = A1 Convincingly shown (no mention of ± scores A0) AG 5 Question Answer Marks Guidance 7(b) d 10d 1 2 x x t x = − 1 1 1 d 10d 2 1 2 x t x   − =   −   1 ln 1 2 10 4 2 x x t B − − − = + M1 A1 Rearrange and attempt to integrate 1 0, 1: 2 t x B = = = − M1 Use initial condition 2 2 40 ln(1 2 ) x t x − = − − − so ( ) 40 2 2 2 1 t x e x e − − = − A1 Convincingly shown, working required AG For large values of t, 1 2 x → B1 CAO 5

This question in 9231/31 Oct/Nov 2020

7 A particle P moving in a straight line has displacement x m from a fixed point O on the line at time t s. -2 200 100 The acceleration of P, in ms , is given by 2 - 3 for x 2 0 . When t = 0 , x = 1 and P has velocity x x -1 10 ms directed towards O. -1 10 ( 1 - 2 x) (a) Show that the velocity v ms of P is given by v = . [5] x … … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that x and t are related by the equation e -40 t = ( 2x - 1) e 2 x -2 and deduce what happens to x as t becomes large. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) 3 2 d 100 200 d v v x x x = − + 2 2 50 200 2 v A x x = − + M1 A1 Correct equation and attempt to integrate Correct 1, 10: 200 x v A = = − = M1 Use initial condition ( ) 2 2 2 100 2 1 x v x − = M1 Rearrange to find 2 v ( ) 10 2 1 x v x − = ± and take negative sign to meet initial condition, so ( ) 10 1 2x v x − = A1 Convincingly shown (no mention of ± scores A0) AG 5 Question Answer Marks Guidance 7(b) d 10d 1 2 x x t x = − 1 1 1 d 10d 2 1 2 x t x   − =   −   1 ln 1 2 10 4 2 x x t B − − − = + M1 A1 Rearrange and attempt to integrate 1 0, 1: 2 t x B = = = − M1 Use initial condition 2 2 40 ln(1 2 ) x t x − = − − − so ( ) 40 2 2 2 1 t x e x e − − = − A1 Convincingly shown, working required AG For large values of t, 1 2 x → B1 CAO 5

This question in 9231/33 Oct/Nov 2020

2 A particle P of mass m kg moves along a horizontal straight line with acceleration ams -2 given by v ( 1 - 2 t 2 ) a = , t where v ms -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t and an arbitrary constant. [3] … … … … … … … … … … … (b) Given that a = 5 when t = 1, find an expression, in terms of m and t, for the horizontal force acting on P at time t. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(a) Separate variables and integrate: 2 1 2   − =     dv t dt v t so 2 ln ln = − + v t t c M1 A1 2 − = t v Ate , 2 − −= t v Ate , 2 − = − t v Ate A1 CAO. 3 Question Answer Marks Guidance 2(b) ( ) ( ) 2 2 2 2 1 2 1 2 − − − − = = − − t t Ate t a Ae t t M1 Substituting their answer to part (a) into given formula ( ) 1, 5 5 t a A e = = = M1 Use initial condition. Force = ( ) 2 1 2 5 2 1 − − t me t A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) 2 1 2 − = v t a t substitute 1, 5 t a = = so 5 = − v M1 Use initial condition. Use N2L, correct work only. Substituting in their answer to part (a) so ( ) 5 = A e M1 Force = ( ) 2 1 2 5 2 1 − − t me t A1 3

This question in 9231/31 Oct/Nov 2021

2 A particle P of mass m kg moves along a horizontal straight line with acceleration ams -2 given by v ( 1 - 2 t 2 ) a = , t where v ms -1 is the velocity of P at time t s. (a) Find an expression for v in terms of t and an arbitrary constant. [3] … … … … … … … … … … … (b) Given that a = 5 when t = 1, find an expression, in terms of m and t, for the horizontal force acting on P at time t. [3] … … … … … … … … … … …

6 marks

Mark scheme: 2(a) Separate variables and integrate: 2 1 2   − =     dv t dt v t so 2 ln ln = − + v t t c M1 A1 2 − = t v Ate , 2 − −= t v Ate , 2 − = − t v Ate A1 CAO. 3 Question Answer Marks Guidance 2(b) ( ) ( ) 2 2 2 2 1 2 1 2 − − − − = = − − t t Ate t a Ae t t M1 Substituting their answer to part (a) into given formula ( ) 1, 5 5 t a A e = = = M1 Use initial condition. Force = ( ) 2 1 2 5 2 1 − − t me t A1 Use N2L, correct work only. Alternative method for question 2(b) ( ) 2 1 2 − = v t a t substitute 1, 5 t a = = so 5 = − v M1 Use initial condition. Use N2L, correct work only. Substituting in their answer to part (a) so ( ) 5 = A e M1 Force = ( ) 2 1 2 5 2 1 − − t me t A1 3

This question in 9231/33 Oct/Nov 2021

6 A particle P moving in a straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The acceleration of P, in ms -2 , is given by 6v v + 9 . When t = 0, x = 2 and v = 72. (a) Find an expression for v in terms of x. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 6 9 dv v v v dx   and attempt to separate variables and integrate 2 9 6 v x A    A1 2, 72 6 x v A    M1 Use initial condition to find constant.   2 9 1 9 v x    A1 Correct, AEF. 4 6(b)     2 9 2 , 9 2 dx dx x x dt dt x x             1 1 1 9 2 2 dx dt x x          M1 Separate variables and write in the form a b dx dt x x c          1 ln 9 2 2 x t B x         A1 Integrate, any correct form. 1 1 0, 2 ln 2 2 t x B    M1 Use initial condition to find constant. 2 18 ln 2 x t x         18 2 2 t x e x   M1 Take logarithms 18 18 2 2 t t e x e   or 18 2 2 1 t x e   A1 Any correct form 5

This question in 9231/31 May/June 2023

6 A particle P moving in a straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The acceleration of P, in ms -2 , is given by 6v v + 9 . When t = 0, x = 2 and v = 72. (a) Find an expression for v in terms of x. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 6 9 dv v v v dx   and attempt to separate variables and integrate 2 9 6 v x A    A1 2, 72 6 x v A    M1 Use initial condition to find constant.   2 9 1 9 v x    A1 Correct, AEF. 4 6(b)     2 9 2 , 9 2 dx dx x x dt dt x x             1 1 1 9 2 2 dx dt x x          M1 Separate variables and write in the form a b dx dt x x c          1 ln 9 2 2 x t B x         A1 Integrate, any correct form. 1 1 0, 2 ln 2 2 t x B    M1 Use initial condition to find constant. 2 18 ln 2 x t x         18 2 2 t x e x   M1 Take logarithms 18 18 2 2 t t e x e   or 18 2 2 1 t x e   A1 Any correct form 5

This question in 9231/32 May/June 2023

6 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P has magnitude 1 2 -t 10 ( 2v - 1) e N and acts towards O. When t = 0 , x = 1 and v = 3 . (a) Find an expression for v in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a)   2 d 1 2 2 1 e d 10    t v v t so   2 d 1 e d 20 2 1    t v t v   e 2 1     t p q A v *M1 Separate variables and attempt to integrate both sides. Where p and q are constants.   1 1 e 2 2 1 20      t A v A1 AEF 3 0, 3, 20          t v A DM1 Substituting the boundary condition and obtain a value. 1 5e 2 3e 1    t t v *M1 A1 Find v in terms of t . AEF. 5 6(b) Integrate:   ln(re )      t x pt q s B *M1   1 5ln(3e 1) 2 3     t x t B A1 AEF 5 0, 1, 1 ln2 3    t x B DM1 Substituting the boundary condition and obtain a value. 1 5 (3e 1) 1 ln 2 3 2    t x t A1 AEF 4

This question in 9231/31 May/June 2024

6 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P has magnitude 1 2 -t 10 ( 2v - 1) e N and acts towards O. When t = 0 , x = 1 and v = 3 . (a) Find an expression for v in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a)   2 d 1 2 2 1 e d 10    t v v t so   2 d 1 e d 20 2 1    t v t v   e 2 1     t p q A v *M1 Separate variables and attempt to integrate both sides. Where p and q are constants.   1 1 e 2 2 1 20      t A v A1 AEF 3 0, 3, 20          t v A DM1 Substituting the boundary condition and obtain a value. 1 5e 2 3e 1    t t v *M1 A1 Find v in terms of t . AEF. 5 6(b) Integrate:   ln(re )      t x pt q s B *M1   1 5ln(3e 1) 2 3     t x t B A1 AEF 5 0, 1, 1 ln2 3    t x B DM1 Substituting the boundary condition and obtain a value. 1 5 (3e 1) 1 ln 2 3 2    t x t A1 AEF 4

This question in 9231/32 May/June 2024

5 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that v 3 - t = x 1 + t and when t = 0, x = 5 . (a) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the magnitude of F when t = 3. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) d x  4  M1 Separate variables, obtain RHS in integrable form. =  − 1  d t x  t + 1  ln x = 4ln t + 1 −+t A A1 t = 0, x = 5: A = ln5 M1 4 t A1 x = 5 ( t + 1) e− 4 5(b) 3 t M1 v = ( 3 − t )  5 ( t + 1) e− dv − t 3 2 3 Acceleration = = 5e − ( t + 1) + ( 3 − t ) 3 ( t + 1) − ( 3 − t )( t + 1) ( ) dt − t 2 AEF Acceleration = 5e ( t + 1) ( 5 − t )(1 − t ) − t 2 M1 F = 2  acceleration, so at F = 10e ( t + 1) ( 5 − t )(1 − t ) At t = 3, magnitude of force is 640 e− N3 A1 31.9 N 3

This question in 9231/31 Oct/Nov 2024

5 A particle P of mass 2 kg moving on a horizontal straight line has displacement x m from a fixed point O on the line and velocity v m s -1 at time t s. The only horizontal force acting on P is a variable force F N which can be expressed as a function of t. It is given that v 3 - t = x 1 + t and when t = 0, x = 5 . (a) Find an expression for x in terms of t. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the magnitude of F when t = 3. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) d x  4  M1 Separate variables, obtain RHS in integrable form. =  − 1  d t x  t + 1  ln x = 4ln t + 1 −+t A A1 t = 0, x = 5: A = ln5 M1 4 t A1 x = 5 ( t + 1) e− 4 5(b) 3 t M1 v = ( 3 − t )  5 ( t + 1) e− dv − t 3 2 3 Acceleration = = 5e − ( t + 1) + ( 3 − t ) 3 ( t + 1) − ( 3 − t )( t + 1) ( ) dt − t 2 AEF Acceleration = 5e ( t + 1) ( 5 − t )(1 − t ) − t 2 M1 F = 2  acceleration, so at F = 10e ( t + 1) ( 5 − t )(1 − t ) At t = 3, magnitude of force is 640 e− N3 A1 31.9 N 3

This question in 9231/33 Oct/Nov 2024

3 A particle P is moving in a straight horizontal line. At time t s, the displacement of P from a fixed point O on the line is x m and the velocity of P is v ms -1 . The acceleration of P is 1 ( v 2 + 4 ) ms -2 in the 2 direction PO. Initially P is at O and is moving with velocity 2 ms -1 . (a) Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the time when P next goes through O. [2] … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) dv 1 2 *M1 Forming differential equation and integrating to a term v + 4 = − 2 ( ) involving an inverse tangent. Condone omission of +A. dt 1 −1 1 1 dv 1 2 to obtain a term v + 4 2 tan ( 2 v ) = − 2 t  + A Allow M1 for integrating = 2 ( ) dt involving an inverse tangent. A1 CWO, condone omission of +A. 1 −1 2 1 DM1 Use initial condition to find constant. t = 0, v = 2 : A = 2 tan ( 2 ) = 8 π dx 1 DM1 Integrate into term involving ln of cosine or sine. v = = 2tan ( 4 π − t ) Accept modulus in ln term. dt 1 OE, for example x = 2ln ( cos ( t − 4 π ) ) + B  1 1 . v = −2tan ( t − 4 π ) , x = 2ln cos ( t − 4 π ) + B x = 0, t = 0: B = ln2 A1 Use initial condition and obtain correct expression. AEF, may see an expression in terms of sec. 1 1  x = 2ln + ln2  = ln cos ( t − 2cos 2 ( t − ( 4 π ) ) ( 4 π ) ) Accept modulus in ln term.   5 3(b) 1 M1 Solve equation to find a value for t. 2ln + ln2 = 0 ( cos ( t − 4 π ) ) t = 12 π A1 CWO 2

This question in 9231/31 Oct/Nov 2025

7 A particle P of mass m kg moving along a rough horizontal table has displacement x m from a fixed point O on the table and velocity v m s -1 at time t s. The particle P is subject to a resistive force of magnitude mgkv N, where k is a positive constant, and a frictional force of magnitude nmg . The particle P is initially at O with speed U m s -1 . 1 kU + n (a) Show that t = ln e o. [4] gk kv + n … … … … … … … … … … … … … … … … … … … … … … … … It is given that U = 10 , k = 0.04 and n = .02 . (b) Find the distance P moves before coming to rest. [4] … … … … … … … … … … … … … … … … … … (c) Find the average speed of P over the period it is moving. [2] … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) dv B1 Must have m. − mgkv − mg = m dt 1 1 *M1 Separate variables and integrate to a natural t = − dv logarithm term. g  kv +  Ignore modulus signs in this part. 1 t = − ln ( kv + ) + C  gk 1 DM1 Use initial condition to find constant OR use When v = u, t = 0: C = ln ( kU + ) correct limits. gk Dependent on having integrated to obtain a natural logarithm term. 1 1 A1 AG, shown convincingly. t = − ln ( kv + ) + ln ( kU + ) Must see an intermediate step before reaching gk gk given result. 1  kU +  Not dependent on B1. t = ln   gk  kv +   4 7(b) 2 dv M1 Three terms in equation of motion, separating − 5 v − 2 = v variables and expressing integrand in a form dx that can be directly integrated. 5 v 5   5  x = − 2  dv = − 2   1 −  dv May be in terms of k ,U and , for example  v + 5   v + 5  1   k  x = − 1 − dv .    gk v + k     Use of suvat in this part cannot be awarded any credit. x = − 52 v + 252 ln ( v + 5 ) + C  M1 Correct form. May be in terms of k ,U and , for example 1    x = − v + ln  v +  + C . 2 gk gk  k  When x = 0, v = 10: C = 25 − 252 ln (15 ) M1 Use initial condition to find constant OR use correct limits in a definite integral. 25 When v = 0, x = 25 − 252 ln (15 ) + 2 ln ( 5 ) = 11.3 [m] A1 Alternative method for question 7(b) 2 M1 Rearrange answer to part 7(a) in the form − t  15  5 − 5 From part (a), t = 52 ln   , v = 15e v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 75 − 52 t M1 Integrate to correct form. x = − 2 e − 5t  + B  75 75 − 52 t 75 M1 Use initial condition to find constant OR use When t = 0, x = 0, B = 2 , x = − 2 e − 5t + 2 correct limits in a definite integral. When v = 0, t = 2.5ln3, x = 25 − 252 ln3 = 11.3 [m] A1 7(b) Alternative method for question 7(b) 2 t M1 Rearrange answer to part 7(a) in the form 5  15  − 5 , v = 15e − 5 From part (a), t = 2 ln   v = ae bt + c .  v + 5  Use of suvat in this part cannot be awarded any credit. 2.5ln3 − 52 t M1 When v = 0, t = 2.5ln3 so x =  0 (15e − 5 ) dt 75 x =  − − 52 t − 5t  2.5ln3 M1 Integrate to correct form. 2 e   0 75 1 75 25 A1 [m] x = − 2 −3 5 12.5  ln3 −−( 2 ) = 25 − 2 ln3 = 11.3 4 7(c) "11.3" "11.3" M1 Divide their x from part 7(b) by the value of t Average speed = = 1  0.04  10 + 0.2  2.5ln3 when v = 0 using the given answer from part ln   7(a). Condone if their x from part 7(b) is 10  0.04  0.2  obtained using suvat. 4.10 [m s–1] A1 4.102, accept values in the range 4.10 – 4.11. 2

This question in 9231/32 Oct/Nov 2025

3 A particle P is moving in a straight horizontal line. At time t s, the displacement of P from a fixed point O on the line is x m and the velocity of P is v ms -1 . The acceleration of P is 1 ( v 2 + 4 ) ms -2 in the 2 direction PO. Initially P is at O and is moving with velocity 2 ms -1 . (a) Find an expression for x in terms of t. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the time when P next goes through O. [2] … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) dv 1 2 *M1 Forming differential equation and integrating to a term v + 4 = − 2 ( ) involving an inverse tangent. Condone omission of +A. dt 1 −1 1 1 dv 1 2 to obtain a term v + 4 2 tan ( 2 v ) = − 2 t  + A Allow M1 for integrating = 2 ( ) dt involving an inverse tangent. A1 CWO, condone omission of +A. 1 −1 2 1 DM1 Use initial condition to find constant. t = 0, v = 2 : A = 2 tan ( 2 ) = 8 π dx 1 DM1 Integrate into term involving ln of cosine or sine. v = = 2tan ( 4 π − t ) Accept modulus in ln term. dt 1 OE, for example x = 2ln ( cos ( t − 4 π ) ) + B  1 1 . v = −2tan ( t − 4 π ) , x = 2ln cos ( t − 4 π ) + B x = 0, t = 0: B = ln2 A1 Use initial condition and obtain correct expression. AEF, may see an expression in terms of sec. 1 1  x = 2ln + ln2  = ln cos ( t − 2cos 2 ( t − ( 4 π ) ) ( 4 π ) ) Accept modulus in ln term.   5 3(b) 1 M1 Solve equation to find a value for t. 2ln + ln2 = 0 ( cos ( t − 4 π ) ) t = 12 π A1 CWO 2

This question in 9231/33 Oct/Nov 2025